//<^? 


IN  MEMORIAM 
FLORIAN  CAJORI 


Digitized  by  the  Internet  Archive 

in  2008  with  funding  from 

IVIicrosoft  Corporation 


http://www.archive.org/details/firstprinciplesoOOslaurich 


FIRST  PRINCIPLES  OF  ALGEBRA 
Complete  Course 


BY 


H.    E.    SLAUGHT,    Ph.D.,  Sc.D. 

ASSOCIATE   PROFESSOR   OF   MATHEMATICS   IN   THE   UNIVERSITY 
OF  CHICAGO 

AND 

N.   J.    LENNES,    Ph.D. 

INSTRUCTOR    IN   MATHEMATICS   IN    COLUMBIA   UNIVERSITY 


Boston 

ALLYN    AND    BACON 

1912 


COPYRIGHT.  1912, 
BY  H.  E,  SLAUGHT 
AND   N.  J.  LENNES. 


Norbiooli  ^xt9fi 

J.  8.  Gushing  Co.  —  Berwick  &  Smith  Co. 

Norwood,  Mass.,  U.S.A. 


^ 


PREFACE 

The  First  Principles  of  Algebra  includes  three  parts,  the 
first  and  second  of  which  constitute  an  Elementary  Course, 
designed  solely  for  the  first  year,  and  the  third  an  Advanced 
Course  for  complete  review  and  further  study.  Each  part  is 
designed  to  cover  the  work  of  one  half  year,  and  the  complete 
book  will  amply  meet  the  entrance  requirements  of  any  college 
or  technical  school. 

The  Elementary  Course  has  two  chief  aims : 

(1)  To  provide  a  gradual  and  natural  introduction  to  the 
symbols  and  processes  of  algebra. 

(2)  To  give  vital  purpose  to  the  study  of  algebra  by  using 
it  to  do  interesting  and  valuable  things. 

Each  of  these  aims  leads  to  the  same  order  of  topics,  which, 
howeverj  differs  somewhat  from  the  conventional  order. 

If  it  is  admitted  that  there  should  he  a  gradually  increasing 
complexity  of  forms  to  be  manipulated,  it  follows  that  factoring 
and  complicated  work  in  fractions  have  no  proper  place  in  the 
first  half  year.  This  book  is  arranged  so  that  factoring  may 
begin  with  the  second  semester  and  complicated  fractions  may 
come  still  later.     Simple  fractions  are  treated  in  Chapter  V. 

The  pupil  is  introduced  to  the  algebraic  notation  by  recalling 
and  stating  in  terms  of  letters  certain  rules  of  arithmetic  with 
which  he  is  already  familiar.     The  simplicity  of  the  algebraic 


jv5i?0fJ^45 


IV  PREFACE 

formulas,  compar6d  witli  the  arithmetical  statement  of  rules 
known  to  the  pupil,  cannot  fail  to  impress  him  with  the  use- 
fulness and  power  of  the  subject  which  he  is  about  to  study. 
This  impression  will  be  deepened  when,  in  Chapter  VI,  rules 
which  caused  considerable  trouble  in  arithmetic  are  derived 
with  the  utmost  ease  by  algebraic  processes. 

The  study  of  equations  and  their  uses  is  naturally  the  main 
topic  of  the  Elementary  Course.  This  topic  is  therefore  devel- 
oped early,  simple  simultaneous  equations  being  completed  in 
Part  One.  It  is  recognized  that  abstract  equations  will  appear 
of  little  or  no  value  to  the  pupil  unless  he  finds  uses  for  them. 
Hence  frequent  lists  of  problems  are  provided  for  translation 
into  equations  and  for  solution. 

Many  of  these  problems  involve  valuable  mathematical  con- 
cepts, such  as  the  dimensions  and  areas  of  rectangles  and  tri- 
angles, the  constitution  of  the  decimal  system  (digit  problems), 
the  fundamental  relations  connecting  velocity,  distance,  and 
time,  properties  of  the  lever,  and  so  forth.  There  are  also 
numerous  artificial  problems  involving  relations  imposed  upon 
abstract  numbers  or  upon  interesting  informational  data. 

In  this  way,  during  the  first  half  year,  algebra  is  made  to 
appeal  to  the  higher  and  more  useful  tyjoes  of  interest,  and  not 
merely  to  the  instinct  for  solving  puzzles,  which  must  be  the 
case  if  the  greater  part  of  this  time  is  spent  on  factoring  and 
in  manipulating  complicated  fractions. 

The  work  of  the  second  half  year  is  intended  to  begin  with 
special  products  and  factors,  long  division  being  introduced  in 
this  connection.  Factoring  is  at  once  applied  to  the  solution 
of  quadratic  equations  and  problems  depending  on  them. 

In  the  chapters  on  radicals  and  quadratic  equations,  simple 
facts  from  geometry,  which  the  pupil  learned  in  arithmetic, 
afford  the  basis  for  a  large  number  of  problems  that  are  of 


PREFACE  V 

legitimate  interest  both  on  their  own  account,  and  because  of 
the  valuable  geometrical  relations  involved. 

The  more  complicated  algebraic  fractions,  which  serve  little 
purpose  in  a  first  course  except  to  afford  drill  in  algebraic 
manipulation,  are  placed  at  about  the  middle  of  Part  Two  and 
are  followed  by  chapters  on  Eatio,  Variation,  and  Proportion, 
Equations  involving  Algebraic  Fractions,  and  a  General  Review. 

For  the  development  of  skill  in  algebraic  manipulation  it 
is  not  sufficient  to  solve  a  certain  number  of  exercises  when  an 
operation  is  first  introduced.  To  fix  each  operation  in  the 
learner's  mind,  there  must  be  recurring  drills  extending  over  a 
considerable  period  of  time.  These  are  amply  provided  for  in 
this  book.  The  fundamental  operations  on  integral  and  simple 
fractional  expressions,  the  solution  of  simple  equations,  and 
the  representation  of  given  conditions  in  algebraic  symbols  are 
constantly  reviewed  in  the  numerous  lists  of  "  drill  exercises," 
many  of  which  may  be  solved  mentally.  Factoring  is  prac- 
tised almost  daily  throughout  the  second  half  year. 

The  principles  of  algebra  used  in  the  Elementary  Course  are 
enunciated  in  a  small  number  of  short  rules  —  eighteen  in  all. 
The  purpose  of  these  rules  is  to  furnish,  in  simple  form,  a  codi- 
fication of  those  operations  of  algebra  which  require  special  em- 
phasis.   Such  a  codification  has  several  important  advantages : 

By  constant  reference  to  these  few  fundamental  statements 
they  become  an  organic,  and  hence  a  permanent,  part  of  the 
learner's  mental  equipment. 

By  their  systematic  use  he  is  made  to  realize  that  the  pro- 
cesses of  algebra,  which  seem  so  multifarious  and  heteroge- 
neous, are,  in  reality,  few  and  simple. 

Such  a  body  of  principles  furnishes  a  ready  means  for  the 
correction  of  erroneous  notions,  a  constant  incitement  to  effec- 
tive review,  and  a  definite  basis  upon  which  to  proceed  at  each 
stage  of  progress. 


VI  PREFACE 

In  the  Advanced.  Course  the  development  is  based  upon  the 
following  important  considerations : 

(1)  The  pupil  has  had  a  one  year's  course  in  algebra,  involv- 
ing constant  application  of  its  elementar}^  processes  to  the 
solution  of  concrete  problems.  This  has  invested  the  pro- 
cesses themselves  with  an  interest  which  now  makes  them  a 
proper  object  of  study  for  their  own  sake. 

(2)  The  pupil  has,  moreover,  developed  in  intellectual  ma- 
turity and  is,  therefore,  able  to  comprehend  processes  of 
reasoning  with  abstract  numbers  which  were  entirely  beyond 
his  reach  in  the  first  year's  course. 

In  consequence  of  these  considerations,  the  treatment 
throughout  is  from  a  more  mature  point  of  view  than  in  the 
Elementary  Course.  Kelatively  greater  space  and  emphasis 
are  given  to  the  manipulation  of  standard  algebraic  forms, 
such  as  the  student  is  likely  to  meet  in  later  work  in  mathe- 
matics and  physics,  and  especially  such  as  were  too  complicated 
for  the  Elementary  Course. 

Attention  is  called  to  the  following  special  features  of  the 
Advanced  Course : 

The  clear  and  simple  treatment  of  equivalent  equations  in 
Chapter  III. 

The  discussion  by  formula,  as  well  as  by  graph,  of  incon- 
sistent and  dependent  systems  of  linear  equations. 

The  unusually  complete  treatment  of  factoring  and  the  clear 
and  simple  exposition  of  the  general  process  of  finding  the 
Highest  Common  Eactor,  in  Chapter  V. 

The  careful  discrimination  in  stating  and  applying  the 
theorems  on  powers  and  roots  in  Chapter  VI. 

The  unique  treatment  of  quadratic  equations  in  Chapter  YII, 
giving  a  lucid  exposition  in  concrete  and  graphical  form  of 
distinct,  coincident,  and  imaginary  roots. 


PREFACE  vii 

The  concise  treatment  of  radical  expressions  in  Chapter  X, 
and  especially — an  innovation  mnch  needed  in  this  connec- 
tion —  the  rich  collection  of  problems,  in  the  solution  of  which 
radicals  are  applied. 

The  authors   gratefully  acknowledge  the  receipt  of   many 

helpful  suggestions  from  teachers  who  have  used  their  High 

School  Algebra. 

H.   E.    SLAUGHT. 

N..  J.   LENNES. 
Chicago  and  New  York,  * 

June,  1912. 


TABLE  OF  CONTENTS 

Elementary  Course 

PART  ONE 

Chapter 

I.  Introduction  to  Algebra 

II.  Equations  and  Problems 

III.  Positive  and  Negative  Numbers    .... 

IV.  Polynomials 

y^  V.  Simple  Fractions    ....... 

VI.     Literal  Equations  and  their  Uses 

VII.     Graphic  Representation 102-115 

VIII.     Simultaneous  Equations 116-133 


Pages 
1-19 

20-35 

36-55 

56-80 

81-91 

92-101 


PART  TWO 

IX.  Products,  Quotients,  and  Factors  , 

X.  Equations  Solved  by  Factoring 

"  XI.  Square  Roots  and  Radicals    . 

XII.  Quadratic  Equations 

XIII.  Algebraic  Fractions 

XIV.  Ratio,  Variation,  and  Proportion 

XV.  Equations  involving  Fractions 

XVI.     General  Review 

vlii 


134-165 
166-175 
176-198 
199-214 
215-235 
236-244 
245-254 
255-27'J» 


TABLE   OF   CONTENTS  ix 

Advanced  Course 

PART  THREE 

Chaptee  Pages 

I.     Fundamental  Laws 277-286 

V  II.     Fundamental  Operations 287-300 

III.  Integral  Equations  of  the  First  Degree  in  One  Un- 

known        301-313 

IV.  Integral  Linear  Equations  —  Two  or  More  Unknowns  314-328 
^^V.     Factoring 329-348 

VI.     Powers  and  Roots 349-364 

VII.     Quadratic  Equations 365-397 

VIII.     Algebraic  Fractions 398-416 

IX.     Ratio,  Variation,  and  Proportion 417-423 

X.     Exponents  and  Radicals 424-448 

XI.     Logarithms 449-456 

XII.     Progressions 457-470 

XIII.     The  Binomial  Formula  ......  471-476 


LIST  OF   PRINCIPLES 


Number 
I. 


II. 

III. 

IV. 

V. 

VI. 

VII. 

VIII. 

IX. 

X. 

XI. 

XII. 

XIII. 

XIV. 

XV. 

XVI. 

XVII. 

XVIII. 


Page 


Addition  and  Subtraction  of  Numbers  having  a  Common 

Factor 

Multiplication  of  the  Sum  or  Difference  of  Two  Numbers 

Division  of  the  Sum  or  Difference  of  Two  Numbers 

Multiplication  of  the  Product  of  Several  Factors 

Division  of  the  Product  of  Several  Factors 

Deducing  One  Equation  from  Another 

Addition  of  Signed  Numbers     . 

Subtraction  of  Signed  Numbers 

Multiplication  of  Signed  Numbers    . 

Division  of  Signed  Numbers 

Arrangement  and  Grouping  of  Terms 

Removal  and  Insertion  of  Parentheses 

Arrangement  and  Grouping  of  Factors 

Multiplication  of  Polynomials    . 

Multiplying  or  Dividing  the  Terms  of  a  Fraction 

Products  of  Powers  of  the  Same  Base 

Quotients  of  Powers  of  the  Same  Base 

Square  Roots  of  Monomials 


10 
12 
13 
24 
39 
43 
47 
49 
67 
62 
67 
70 
82 
136 
138 
177 


FIRST  PRmCIPLES  OF  ALGEBRA 

ELEMENTARY  COURSE 

PART  ONE 

CHAPTER   I 

INTRODUCTION   TO   ALGEBRA 

LETTERS  USED  TO  REPRESENT  NUMBERS 

1.  Algebra,  like  arithmetic,  deals  with  numbers.  In  arith- 
metic numbers  are  represented  by  means  of  the  Arabic  numerals 
0,  1,  2,  3,  4,  5,  6,  7,  8,  9.  In  algebra  letters,  as  well  as  these 
numerals,  are  used  to  represent  numbers. 

2.  One  advantage^  in  using  letters  to  represent  numbers  is  in 
abbreviating  the  rules  of  arithmetic. 

Examples.  What  is  the  area  of  a  blackboard  whose  length  is 
12  feet  and  whose  width  is  3  feet  ?  What  is  the  area  of  a  city 
lot  which  is  25  feet  wide  and  120  feet  deep  ? 

In  arithmetic  we  multiply  the  length  by  the  width  and  the  product 
is  the  area.  In  algebra  we  abbreviate  by  letting  I  represent  the  length, 
w  the  jvidth,  and  a  the  area. 

The  rule  then  stands  as  follows: 
/  xw  =  a 
This  is  called  a  formula  and  covers  all  possible  cases. 

It  is  to  be  noted  that  /  and  w  represent  the  number  of  units  in 
the  length  and  width  respectively  and  a  the  number  of  square 
units  in  the  area. 

1 


2  INTRODUCTION  TO  ALGEBRA 

3.  The  signs  of  operation,  +,  — ,  x,  -^,  and  the  sign  of 
equality,  =,  are  used  in  algebra  with  the  same  meaning  that 
they  have  in  arithmetic.  However,  the  sign  of  multiplication 
is  usually  omitted  between  letters,  and  between  an  Arabic 
figure  and  a  letter;  but  a  sign  omitted  between  two  Arabic 
figures  always  means  addition.  Multiplication  is  also  indicated 
by  a  point  written  above  the  line. 

Thus,  2  X  I  X  w  may  be  written  2  ■  I .  w  or  2  Iw. 
But  25  means  20  +  5,  not  2  •  5. 

Division  may  be  indicated  by  a  fraction,  as  in  arithmetic. 

2  a 

Thus,  2  -f-  3  is  written  -  and  a  -f-  &  is  written  -  • 
o  0 

Historical  Note.  The  Arabs  brought  our  present  system  of  numerals 
to  Europe  when  they  invaded  Spain  in  the  eighth  century  of  our  era.  It 
is  now  known  that  these  numerals  (including  tlie  zero)  are  really  of  Hindu 
origin,  and  that  they  were  invented  between  the  years  200  a,d.  and  650  a.d. 
Of  all  mathematical  inventions  no  one  has  contributed  more  to  general  in- 
telligence or  to  practical  welfare  than  this  one.  To  appreciate  this,  one  has 
only  to  try  to  multiply  two  numbers  such  as  589  and  642  when  expressed 
in  the  Roman  notation  ;  that  is,  to  multiply  DLXXXIX  by  DCXLII. 

It  was  about  the  year  1500  a.d.  that  our  present  symbols  indicating  addi- 
tion and  subtraction  first  appeared  in  a  book  by  a  German  named  Johann 
Widemann.  The  sign  x  for  multiplication  was  first  used  about  50  years 
later  by  an  Englishman,  William  Oughtred.  About  the  same  time  the  sign 
= was  first  used  by  Recorde,  also  an  Englishman  ;  but  the  sign  -r-for  division 
does  not  appear  until  1659  when  it  was  used  by  a  German,  J.  H.  Rohn. 

4.  In  the  following  exercises,  note  the  simplicity  with  which 
the  rules  of  arithmetic  may  be  stated  by  means  of  letters.  The 
systematic  use  of  letters  to  represent  numbers  is  one  of  the 
chief  points  of  difference  between  algebra  and  arithmetic. 

EXERCISES 

1.  A  box  is  6  inches  long,  4  inches  wide,  and  3  inches  high. 
How  many  cubic  inches  does  it  hold  ? 

2.  How  many  cubic  feet  of  air  in  a  schoolroom  which  is  35 
feet  long,  25  feet  wide,  and  15  feet  high  ? 


What  is  the 
a      ,  c 


LETTERS   USED   TO   REPRESENT   NUMBERS  3 

3.  Given  the  length,  width,  and  height  of  a  rectangular 
solid,  how  do  you  find  its  volume  ?  State  this  rule  as  a  formula, 
using  I,  w,  h,  and  v  for  the  number  of  units  in  the  length, 
width,  height,  and  volume,  respectively. 

2  5.2-5 

4.  By  a  rule  of  arithmetic  the  product  of  ^  and  -  is  - — -  • 

What  is  the  rule  for  finding  the  product  of  any  two  fractions  ? 
This  rule  may  be  stated  as  a  formula  as  follows : 

a     c__ac 
b     d     bd 

OL  C 

in  which  z-  and  -  are  any  two  fractions. 

,      ^     ..1.      ..  3     5     3^7     3-7 

5.  By  a  rule  of  arithmetic  7-5-;;  =  tX-  =  - — - , 

-^  4     7     4     5     4-5 

rule  for  dividing  one  fraction  by  another  ?     Using  ^  and  -,  to 
represent  any  two  fractions,  state  this  rule  as  a  formula. 

6.  By  arithmetic  8  per  cent  of  90  means  90  x  .08  =  7.2, 
in  which  90  is  called  the  base,  .08  the  rate,  and  7.2  the 
percentage. 

In  like  manner  find  6  per  cent  of  125 ;  5  per  cent  of  350 ;  3 
per  cent  of  80 ;  and  10  per  cent  of  4.9.  In  each  case  how  is  the 
percentage  found  ?     State  the  rule. 

State  this  rule  as  a  formula,  letting  b  represent  the  base,  r  the 
rate  and  p  the  percentage. 

7.  By  a  rule  of  arithmetic  the  interest  on  $  500  for  3  years 
at  6  per  cent  is  500  x  .06  x  3  =  9,  in  which  500  is  the  principal, 
.06  the  rate,  and  3  the  time. 

8.  Find  the  interest  on  $  900  for  7  years  at  4  per  cent,  and 
on  $  1250  for  5  years  at  5  per  cent. 

9.  Given  any  principal,  rate,  and  time,  how  do  you  find  the 
interest  ? 

State  this  rulers  a  formula,  using  p  for  principal,  r  for  rate, 
t  for  time  and  i  for  interest. 


4  INTRODUCTION  TO   ALGEBRA 

ALGEBRAIC  OPERATIONS 

6.  In  algebra  numbers  are  added,  subtracted,  multiplied,  and 
divided  as  in  arithmetic.  There  are,  however,  certain  ways 
of  carrying  out  these  operations  which  are  essential  in  algebra 
though  not  used  extensively  in  arithmetic.  Some  of  these  we 
now  proceed  to  study. 

6.    Definitions.     Any  combination  of  Arabic  numerals,  letters, 

and  signs  of  operation,  used  for  the  purpose  of  representing 

numbers,  is  called  a  number  expression. 

3a  -  2 
E.g.  38,  18  r,  5  n  +  8  n,  7 —  are  number  expressions. 

3^ 2 

The  expression  — 7 —  is  said  to  be  written  in  symbols,  that 

is,  by  means  of  numerals,  letters,  and  signs  of  operation. 

This  number  expression  written  in  words  would  be  "  three  times 
the  number  a  diminished  by  2  and  the  result  divided  by  &." 

EXERCISES 

1.  If  a  and  6  are  numbers,  express  in  symbols  their  sum 
and  also  their  product. 

2.  If  m  and  71  are  numbers,  write  in  symbols  m  divided  by 
n ;  also  m  minus  n. 

3.  If  p  and  q  are  numbers,  write  the  sum  of  5  times  p  and  3 
times  q. 

4.  If  a,  h,  and  c  are  numbers,  write  in  symbols  that  a  multi- 
plied by  h  equals  c ;  also  that  c  divided  by  a  equals  h. 

If  a  =  3,  6  =  5,  c  =  7,  find  the  value  of  each  of  the  following 
number  expressions : 

5.  a  +  h  +  c.  10.    6c  — 3a.  13.  3a6c  — 2  6-f-3c. 

6.  a-\.h-c.  ^^    h±c^  ^^    2a  +  ^h  ^  ^^ 

7.  2a  —  6  +  c.  a  c 

8.  ah -2  c.  Aa-h  3c-2a 

9.  16  a -he.  '        c      '  Sh       ' 


ADDITION   AND   SUBTRACTION  .  5 

7.  Definition.  If  a  number  is  the  product  of  two  or  more 
numbers,  these  are  together  called  the  cofactors  of  the  given 
number,  and  any  one  of  them  is  called  a  factor. 

E.g.     3  and  4  are  cofactors  of  12,  as  are  also  2  and  6,  1  and  12. 
1,  2,  3,  4,  6,  and  12  are  factors  of  12.     a,  h,  c  are  factors  of  abc. 

8.  Definition.  If  a  number  is  the  product  of  two  factors, 
then  either  of  these  factors  is  called  the  coefficient  of  the  other 
in  that  product. 

E.g.  In  2  .  3,  2  is  the  coefficient  of  3,  and  3  is  the  coefficient  of  2. 
In  9  rt,  9  is  the  coefficient  of  rt,  r  is  the  coefficient  of  9  t,  and  t  is  the 
coefficient  of  9  r. 

In  such  expressions  as  9rf  the  factor  represented  by  Arabic 
figures  is  usually  regarded  as  the  coefficient. 

ADDITION    AND    SUBTRACTION    OF    NUMBERS    HAVING    A    COMMON 

FACTOR 

9.  It  is  sometimes  necessary  to  add  number  expressions  like 
5  n  and  8  n  without  first  assigning  a  definite  value  to  n. 

It  seems  clear  that  5  times  any  number  and  8  times  the  same 
number  make  13  times  that  number.     That  is, 
5  ?i  +  8  n  =  13  w. 

Determine  whether  5?iH-8?i  =  13 ?i  when  ti  =  3;  also  when 
w  =  9 ;  when  n  =  12.     Test  this  for  still  other  values  of  n. 

Note  that  the  sum  of  5  •  3  and  8  •  3  may  be  found  by  adding 
the  coefficients  of  the  common  factor  3,  while  the  sum  of  5  n 
and  8  n  must  be  found  in  this  way. 

In  this  manner  perform  the  following  additions : 

1.  ^x-\-lx^lQ,x-ir2x.  4.   764-86  +  &  +  66. 

2.  13?i-|-8n4-T7i  +  9?i.  5.    ^t+lt  +  ^t. 

3.  3a  +  a+2a  +  6a.  6.   3r  +  5r  +  llr. 

Test  the  correctness  of  the  result  in  each  of  the  above  by 
letting  x  =  2,  n  =  l,  a  =  4,  6  =  3,  ^  =  6,  and  r  =  ^. 


6  .  INTRODUCTION  TO   ALGEBRA 

10.  Numbers  having  a  common  factor  may  be  subtracted  in 
a  similar  manner. 

Tbus,  from  64  =  8  •  8  From  84  =  12  •  7  From  17  n 

subtract       48  =  6  •  8      subtract  49  =    7-7        subtract    6  n 
Remainder  16  =  2  •  8  35  =    5  •  7-  lln 

In  this  way  perform  the  following  subtractions : 

1.  8.7-3-7.  5.   10&-46.  9.  14a-8a. 

2.  6.99-5-99.  6.     7a-4a.  10.  126-96. 

3.  6n-2n.  7.   23a;-16a;.  11.  8^-2^. 

4.  8  a  — 3  a.  8.    15  7i  — 3n.  12.  19r  — llr. 

Test  these  results  by  giving  the  same  values  to  the  letters 
as  were  used  on  page  5.     Try  also  other  values. 

It  is  evident  that  any  two  numbers  having  a  common  factor 
may  be  added  or  subtracted  in  this  manner. 

These  examples  illustrate 

Principle  I 

11.  Rule.  To  find  the  sum  or  difference  of  two  numbers 
having  a  common  factor,  add  or  subtract  the  coefficients 
of  tJw  common  factor  and  multiply  the  result  by  the 
common  factor. 

12.  Definition.  The  substitution  of  special  values  for  letters  in 
a  number  expression  in  order  to  test  the  correctness  of  an  opera- 
tion is  called  a  check. 

EXERCISES 

Perform  the  following  indicated  operations  and  check  the 
results  in  the  first  eight  by  assigning  values  to  the  letters : 

1.  68^-11^.  5.  20n-6n  +  2w. 

2.  15n  +  25n-18n.  6.  5^  +  20^-3^. 

3.  10x-15x-irlx-2Sx.  7.  8s-3s-f20s. 

4.  18A;-3A:-2fc  +  6A:.  8.  6a-4a  +  3a-2a. 


ADDITION  AND   SUBTRACTION  7 

9.  7  ab-3ab-\-2ab.  17.  32 ac  -  17  ac  +  2  ac. 

10.  11  rs  -  2  rs  -  5  rs.  18.  91  a  -  81  a  +  2  a. 

11.  34  a;?/ -  18  x?/ -  4  3^2^.  19.  16  .x- +  24  ic -f  8  a;  -  40  a;. 

12.  12  ?/i/i  —  G m?i  —  3mri.  20.  o y  -\- 31y  —  9y  —  21y. 

13.  17.s-^  +  3^'^-12si.  21.  63c-47c-8c4-7c. 

14.  12  a6c  -  2  a6c- 6  a&c.  22.  16  f  -  11 «  -  2  «  +  3  i.  ' 

15.  A2xy-j-6xy-35xy.  23.  12  a;y  -  9  a;?/ +  8  xy. 

16.  29rs^-18/vj^-6rs^.  24.  39  a6  -  27  aft  -  8  a6. 

13.  Definitions.  Two  number  expressions  representing  the 
same  number,  when  connected  by  the  sign  =,  form  an  equality. 

The  expressions  thus  connected  are  called  the  members  of 
the  equality  and  are  distinguished  as  the  right  and  left  members. 

Equalities,  such  as  5n  +  Sn  =  13 n,  in  which  the  letters 
may  be  any  numbers  ichatever,  are  called  identities.  Not  all 
equalities  are  of  this  kind.  For  example,  ti  +  3  =  5  is  true 
only  when  n  =  2. 

When  it  is  desired  to  emphasize  that  an  equality  is  an 
identity,  the  sign  =  is  used.     That  is,  3  a  +  5  a  =  8  a. 

14.  Definitions.  A  parenthesis  is  used  to  indicate  that  some 
operation  is  to  be  extended  over  the  whole  number  expression 
inclosed  by  it.  Thus  2(x  -f  y)  means  that  the  sum  of  x  and  y 
is  to  be  multiplied  by  2,  while  2  x-\-y  means  that  x  alone  is  to 
be  multiplied  by  2. 

Instead  of  a  parenthesis  a  bracket  [  ],  or  a  brace  \  \,  may 
be  used  with  the  same  meaning.  Any  such  symbols  are  called 
symbols  of  aggregation. 

E.g.     2(x  -\-  y),2[x  +  ?/]or  2{x  +  y}  all  mean  the  same  thing. 

Historical  Note.  The  parenthesis  (  )  was  first  used  with  its  present 
meaning  by  an  Englishman,  A.  Girard,  in  a  book  on  "  Arithmetic,"  pub- 
lished in  the  year  1029.  The  bracket  and  brace  are  of  later  origin,  as  is 
also  the  sign  =  to  denote  identity. 


8  INTRODUCTION  TO  ALGEBRA 

MULTIPLICATION  OF   THE  SUM  OR  DIFFERENCE  OF  TWO  NUMBERS 

15.  The  sum  of  two  or  more  numbers  may  be  multiplied  by 
another  number  in  two  ways,  as  shown  in  the  following 
examples : 

(1)  4(2  +  7)-  4.9  =  36, 

ol-  4(2  H-  7)  =  (4  •  2)  +  (4  •  7)  =  8  +  28  =  36. 

(2)  3(3 +  8  + 9)  =  (3.  20)  =  60, 

or       3(3  +  8  +  9)  =  (3  .  3)  +  (3  •  8)  +  (3  .  9)  =  9  +  24  +  27  =  60. 

In  each  case  the  same  result  is  obtained  whether  lue  first  add 
the  numbers  in  the  parenthesis  and  then  multiply  the  sum,  or  first 
multiply  these  numbers  one  by  one  and  then  add  the  products. 

In  case  the  numbers  are  represented  by  letters,  the  second 
process  only  is  available. 

E.g.     3(a  +  &)=3a  +  3&  and  m(r  +  s)  =  mr  -f  ms. 

Multiply  each  of  the  following  in  two  ways  where  possible: 

1.  3(2  +  7).  4.   3(a-f-6).  7.    x(3  +  7  +  10). 

2.  5(3  +  4  +  5).  5.    ll(/i  +  A:).  8.    16{x-\-y  +  z). 

3.  8(5  +  9  +  7).  6.   4(a  +  6  +  c).  9.    20(m  +  ?i+p). 

16.  The  difference  of  two  numbers  in  Arabic  figures  may 
likewise  be  multiplied  by  a  given  number  in  either  of  two  ways. 

E.g.  6(8  -  3)  =  6  ■  5  =  30, 

or  6(8  -  3)  =  (6  .  8)  -  (6  .  3)  =  48  -  18  =  30. 

In  the  case  of  numbers  represented  by  letters  evidently  the 
second  process  only  is  available. 

E.g.  6(r  -  0  =  6  r  -  6  ^  and  a{c  -  d)  =  ac  -  ad. 

Perform  as  many  as  possible  of  the  following  multiplica- 
tions in  two  ways: 

1.7(9-2).         4.17(18-11).  7.  5(a;-l).       10.  m(r-s). 

2.  12(17-7).      5.  9(a-2).  8.  3(2/ -2).        11.  x(y-z). 

3.  5(12-8).       6.  8(^-4).  9.  a{c-d).       12.  t(u-^v). 


MULTIPLICATION  9 

The  foregoing  examples  illustrate 

Principle  II 

17.  Rule.  To  i7vidtiply  the  sinn  or  difference  of  two 
numbers  hy  a  given  number,  multiply  each  of  tJie  num- 
bers separately  by  the  given  number,  and  add  or  subtract 
tlie  products, 

EXERCISES 

1.  Multiply  5+7  +  11  by  3  without  first  adding,  and  then 
check  by  performing  the  addition  before  multiplying. 

2.  Multiply  m  +  n  by  4  and  check  for  m  =  5,  ?i  =  7. 

4(m  +  n)=  4;^  +  4  u 
Check,  4(5  4-  7)  =  4  .  12  =  48,  also 

4(5  +  7)  =  (4  .  5)  +  (4  •  7)  =  20  +  28  =  48. 

3.  Multiply  a;  +  2/  by  r  and  check  for  a;  =  2,  ?/  =  4,  r  =  6. 

4.  Multiply  r  +  .s  by  A:  and  check  for  r  =  4,  s  =  5,  A:  =  6. 

5.  Multiply  a  +  6  +  c  by  m  and  check  for  a  =  3,  6  =  2,  c  =  1, 
m  =  4.  • 

6.  Multiply  m  —  ?i  +  2  by  c  and  check  for  ?7i  =  5,  7i  =  2  and 
c  =  3. 

7.  Multiply  a  —  6  —  c  by  d  and  check  for  a  =  10,  6  =  3,  c  =  4 
and  d  =  8. 

8.  Multiply  r  +  s  +  9  — «  by  c  and  check  for  r  =  1,  s  =  8, 
t  =  3,  and  c  =  2. 

Find  the  following  products: 

9.  8(13-5).  16.    5(7  + a- 6). 

10.  2(12+41-36).  17.  3(x  +  2-?/). 

11.  9(a  +  8-&).  '  18.  8(«  +  6-c  +  d).  ' 

12.  3(a  +  x  +  2/-l).  19.  8(9 -a +6 +3). 

13.  3(a  +  ft-c).  20.  a(3  +  6-c). 

14.  a(18  — 7).  21.  c(m-n+p). 

15.  7(3  +  8  + 9 -a).  22.  19(8  -  a:  -  y  +  z). 


(2)  (20  -  12) 

or  (20  -  12) 


10  INTRODUCTION  TO   ALGEBRA 

DIVISION   OF  THE   SUM  OR  DIFFERENCE  OF  TWO  NUMBERS 

18.  In  dividing  the  sum  or  difference  of  two  numbers  hj  a 
given  number,  when  these  are  represented  by  Arabic  figures, 
the  process  may  be  carried  out  in  two  ways.     Thus, 

(1)  (12  +  8)^2  =  20^2  =  10, 

or         *  (12  +  8)  -  2  =  (12  -  2)  +  (8  -  2)  =  6  +  4  :..  10. 

4  =  8-4  =2, 

4  =  (20  -  4)  -  (12  ^  4)  =  5  -  3  =  2. 

Describe  each  of  these  two  ways  of  dividing  a  sum  or  differ- 
ence by  a  number.     How  do  the  results  compare  ? 

If  the  numbers  in  the  dividend  are  represented  by  letters, 
the  division  can  usually  be  carried  out  only  in  the  second 
manner  shown  above. 

E.g.     (r+0-^5=(r-^5)  +  (/-^5),or,  ^=^4-|. 

In  either  case  this  is  read:  r  plus  t  divided  by  5  equals  r 
divided  by  5  2^lus  t  divided  by  5. 

In  this  manner  perform  each  of  the  following  divisions  in 
two  ways  when  possible  and  check  the  results : 

1.  (16  +  12)-^4.  3.    (a;-i/)--6.         5.    (m-?')H-a. 

2.  (20 -10) -5.         4.    {x-\-z)~3.         6.    (m  +  r)--&. 

These  examples  illustrate 

Principle  III 

19.  Rule.  To  divide  the  sum  or  difference  of  two 
numbers  by  a  given  number  divide  each  number  sepa- 
rately and  add  or  subtract  the  quotients. 

EXERCISES 

1.  Divide  72  +  56  by  8  without  first  adding. 

2.  Divide  144  —  36  by  12  without  first  subtracting. 

3.  Divide  r  -\- 1  by  5  and  check  the  quotient  when  r  =  15, 
f  =  25 ;  also  when  r  =  60,  i  =  75, 


DIVISION  11 

•  4.    Multiply  7  +  9  by  3  without  first  adding  7  and  9. 

5.  Multiply  25  —  8  by  5  without  first  subtracting. 

6.  Find  the  product  of  12  and  a  +  h,  checking  the  result 
when  a  =  5,  b  =  7. 

Perform  the  following  indicated  operations : 

7.  3(a  +  &  +  c  +  d).     Check  for  a  =  1,  6  =  2,  c  =  3,  fZ  =  4. 

8.  'K^r—s-^-t  —  x).      Check  for  r  =  ^  =  5,  s  =  a;  =  4. 

9.  (m  +  w  +  r)  -j-  4.      Check  for  m  =  64,  n  =  32,  r  =  8. 

10.  (x  -{- y -\- z) -i- 5.       Check  for  x  =  100,  y  =  50,  and  z  =  25. 

11.  74rs-67rs-2rs-3rs.         13.    a(4-dH-6+c+3). 

12.  (63-35-14+21) -7.  14.    (21- x-y+S+c)k. 

15.  A9pq -\- IS  2)q  — 62 pq-{-Spq. 

16.  IS  xyz -\- S  xyz  —  S  xyz  —  7  xyz. 

17.  ((/-hr  +  s  +  «  — a  — 6)-=-c. 

18.  35 Im  —  33lm-{-7l7n  —  3  Im  —  2 Zm. 

19.  A:(Z  +  ?/i  +  n  +  r  — s  — <). 

20.  a(c-\-d-c-\-f—g). 

21.  27  a6c  —  19  aZ>c  —  4  a6c' -f  8  a6c. 

22.  (a-[-r-\-s-t  —  q)-i-S, 

23.  (12-a;  +  ?/-2)^6. 

24.  For  what  values  of  a,  6,  c,  d  are  the  following  equalities 
t^^e  ?  (^)  «^  ^  (j(.  +  a(Z  =  a(6  -f  c  +  d). 

(&)  a6  +  ac  —  ad  =  a(b  -\-  c  —  d). 

Historical  Note.  The  fundamental  character  of  Principles  II  and  III 
was  not  fully  appreciated  until  the  first  part  of  the  last  century.  Principle 
II  states  what  is  called  the  Distributive  Law  of  Multiplication  with  re- 
spect to  addition  and  subtraction.  That  is,  the  multiplier  is  distributed 
over  the  multiplicand.  The  name  was  first  used  by  a  Frenchman  F.  J. 
Servois,  in  a  paper  published  in  1814.  Principle  III  states  the  same  law 
for  division.    Compare  notes  on  pages  57,  67. 


12  INTRODUCTION  TO  ALGEBRA 

MULTIPLICATION   OF   A  PRODUCT  ' 

20.  In  arithmetic  to  multiply  a  product  like  3  •  5  by  2,  we 
first  multiply  3  by  5  and  this  result  by  2. 

Thus,  2.  (3- 5)  =2.15  =  30. 

We  cannot,  however,  multiply  3  ?i  by  2  in  this  way.  But  it 
is  obvious  that  twice  3  7i  is  3  w  +  3  n  =  6  7i. 

Thus,  if  n  =  5,  2  •  (3  •  5)  =  6  •  5. 

Hence  in  multiplying  the  product  3  •  5  by  2  we  may  multiply 
the  factor  3  only.  Would  you  get  the  same  result  if  you 
multiplied  the  factor  5  only  ? 

Example.  Find  the  product  2(3  •  4  •  5)  in  as  many  ways  as 
possible. 

Solution : 

2(3  .  4  .  5)=  2  .  60       =  120,    and  2(3  •  4  •  5)  =  3  •  8  •  5    =  120, 
Also  2(3  .  4  .  5)  =  6  .  4  •  5    =  120,    and  2(3  •  4  •  5)  =  3  •  4  •  10  =  120. 

In  like  manner  find  the  following  products  : 
1.   5(3  .  7).     2.   8(2  .3.4).      3.   9(2  •  5).  4.    Q>{5  .  3). 

5.    3(5  .  2).     6.    4(7  .  2  .  3).      7.   7(2  .  4  .  3).      8.    5(2  .  5  •  3). 
These  examples  illustrate 

Principle   IV 

21.  Rule.  To  Tnultiply  tJie  produvt  of  several  factors 
hy  a  ^iven  number,  jnultiply  any  one  of  the  factors  hy  that 
number^ 

22.  Principles  IV  and  II  should  be  carefully  contrasted,  as 
in  the  following  example  : 

2  (2. 3.  5)  =4. 3- 5  =  2. 6. 5  =  2. 3. 10, 

but  2(2  +  3  +  5)=44-^6-f-10. 

In  multiplying  the  product  of  several  numbers  we  oiierate  upon 
any  one  of  them,  hut  in  multiplying  the  sum  or  difference  of  num- 
bers we  operate  upon  each  of  them. 


DIVISION  OF   A  PRODUCT  13 

EXERCISES 

Multiply  as  many  as  possible  of  the  following  in  two  or 
more  ways.     Check  where  letters  are  involved. 

1.  7(3.4.5).  5.   S(5xy).  9.   15(7 ah). 

2.  8(7.2.3).  6.   S{Sxyz).  10.   3(4m7i). 

3.  9(2.3.4).  7.   4(19.25).  11.    7  {2  xy). 

4.  5(2a6).  *  8.   5(7  a6c).  12.   2(16r8). 

DIVISION  OF   A  PRODUCT 

23.  Division  of  the  product  of  several  factors  by  a  given 
number  may  be  performed  in  various  ways : 

E.g.  (4.6.10)-j-2  =  240-f-2  =  120. 

Also  (4.6.10)-^2  =  2.6.10  =  120, 

(4.6.10)--2  =  4.3.10  =  120, 
and  (4  .  6  .  10)^  2  =  4  .  6  .  5  =  120. 

Note  that  in  each  case  only  one  factor  is  divided. 
Perform  each  of  the  following  divisions  in  more  than  one 
way  where  possible: 

1.  (5.  8- 3)- 2.     4.    (11.  20. 16) -4.     7.    (10.35-3)^5. 

2.  20  abc  -7-4.         5.    14  xyz  -i-  7.  8.    14  xyz  -5-  x. 

3.  12«&c-^3.         6.    12  aftc -5- c.  9.    (12-40.13)^8. 

These  examples  illustrate 

Principle  V 

24.  Rule.  To  divide  the  product  of  several  factors 
by  a  given  niumber  divide  any  one  of  the  factors  hy  that 
nuinber. 

25.  Principle  V  is  already  known  in  arithmetic  in  the  process 
called  cancellation. 

Thus,  in  the  fraction  "  \  '  '  ,  3  may  be  canceled  out  of  either  6  or 
9,  giving  "  •  ^'^  =  2  .  2  .  9  or  2  .  6  . 3. 


14  INTRODUCTION   TO   ALGEBRA 

EXERCISES 

1.  Contrast  Principles  III  and  V. 

By  Principle  Y,  (4.6.8)-2:iz2.6.8  =  4.3.8  =  4.6.4. 
By  Principle  HI,  (4  +  6  +  8)  -f-  2  =  2  +  3  +  4. 

That  is  in  dividing  the  product  of  several  nmnhers  by  another 
number  ive  operate  upon  any  one  of  them,  but  in  dividing  their 
sum  or  difference  we  operate  upon  each  of  them. 

Perform  the  following  indicated  operations  : 

2.  5  abc  -t-a.  9.  3  a  (7  —  c  +  5)  -;-  a. 

3.  15xy-r-S.  10.  5  6c(d— e+3)  ^6c. 

4.  5(2x-hSy).  11.  19(3a-66-h9c)--3. 

5.  (33  aj- 44?/) --11.  12.  13(8  -  4  6  +  12  a) -4. 

6.  (78  s -39  0-^13.  13.  14(7  -  7  m  +  14n) -j- 7. 

7.  8(3a-f-2?/).  14.  12a(36-3c  +  9)^3. 

8.  15(a-6  +  c) --5.  15.  24(16  6  -  8  c  +  24  cZ) --8. 

16.  Divide  7  a  •  14  6  •  21  c  by  7  in  three  different  ways. 

17.  Add  5  a,  ^^,  ?^,  and  i|^,  using  Principles  V  and  I. 

^  o  o 

18.  From  ?^  subtract  ?1M^  using  Principles  V  and  I. 

^^    ^:^         14  a  ,  10  a      ,  ,       ,  6  a 

19.  From  —^-\-—z-  subtract  -— • 

20.  Find  the  sum  of  i|^,  ?^,  i^",  7  x,  and  3  x. 

8         5         4 

«-.     -I7-   ^  ^v.  f  100  rs    90  rs    ^    .  25  rs 

21.  Find  the  sum  of     ^^    ,      ^    ,  and  » 

10  9  o 


22.    From  25  a;?/  subtract  — ^-' 


.  ,  ^  8  abc  ,  18  a6  ,  7  a&cZ  ,  a6e 

23.  Add  +  -^— +  — ^  +  -'^' 

c  3  a  e 

A  J  1  18  a  ,  5  6a  ,  4  a;a  ,  a??i 

24.  Add  ——  -f  — —  H h  —  • 

b  b  x         m 


ORDER   OF   INDICATED    OPERATIONS  15 

ORDER  OF  INDICATED  OPERATIONS 

26.  In  a  succession  of  indicated  operations  the  final  result  de- 
pends in  some  cases  upon  the  oj-der  in  which  the  steps  are  taken : 

Thus,  6  +  3-8  would  give  9  •  8  =  72,  if  the  addition  were  performed 
first ;  and  would  give  6  +  24  =  30,  if  the  multiplication  were  per- 
formed first. 

Similarly,  24  ^  2  •  3  would  give  12  .  3  =  36,  if  the  division  were  per- 
formed first;  and  would  give  24  -=-  6  =  4,  if 'the  multiplication  were 
performed  first. 

27.  However,  when  only  additions  and  subtractions  are  involved 
and  no  symbols  of  aggregation  occur,  the  result  is  the  same  no 
matter  in  what  order  the  operations  are  performed. 

Test  this  statement  by  performing  each  of  the  following  in- 
dicated operations  in  several  different  orders : 

1.  8  -  3  -h  14  -  4.  3.   7  -  6  +  1  +  3. 

2.  13-1-4-3  H- 2.  4.    15-12  +  4-2. 

A  similar  statement  holds  in  some  cases  when  only  multipli' 
cations  and  divisions  are  involved. 

Thus,  12  X  6  ^  3  =  24  no  matter  in  which  order  the  operations  are 
performed,  while  24-4-2-3  may  equal  4  or  36. 

In  case  of  doubt  symbols  of  aggregation  should  be  used  to 
show  the  order  intended. 

Thus,  24  -e-  (2  .  3)  =  4,  while  (24  -  2)  •  3  =  36. 

And  18  -f-  (6  -4-  2)  =6,  while  (18  ^  6)  -f-  2  =  f. 

The  following  rule  is  in  accordance  with  universal  custom : 

28.  In  an  expression  involving  additions,  subtractions,  multi- 
plications, and  divisions  without  any  symbols  of  aggregation,  all 
midtiplications  and  divisions  are  to  be  performed  before  any  addi- 
tions or  subtractions. 

E.g.  54-3.4-8--2  =  5-M2-4  =  13,  and  8-2-4-2-2x2 
+  3  =  8-l-4-f3  =  6. 


16  INTRODUCTION   TO   ALGEBRA 

EXERCISES 

Perform  the  following  indicated  operations: 

1.  20  +  5-3  +  4-2.  5.    (28 --4). 2 +  10. 

2.  12^3-2  +  5.  6.   5.3  +  2.7-19. 

3.  (28.4)-2  +  10.  7.   26-4-2-6.2. 

4.  28(4--2)  +  10.   .  8.    18-2.6-8--4  +  20. 

♦  DRILL  EXERCISES 

Note.  —  The  pages  headed  "  Drill  Exercises  "  are  intended  to  afford 
general  practice  in  performing  algebraic  operations.  Each  such  page 
consists  of  miscellaneous  exercises  on  the  various  topics  that  precede. 

If  a  =  5,  6  =  3,  c  =  2j  find  the  numerical  value  of  each  of  the 
following : 

1.  ac  +  bc  and  (a  +  b)c.  5.    a(b  —  c)  and  ab  —  ac. 

2.  ac  —  be  and  (a  —  b)c.  6.   — ,  -  x  6  and  a  X  - 

3.  abc,  ax  (be)  and  b x  (ac). 

4.  a(b  +  c)  and  ab  +  ac. 

Perform  the  following  indicated  operations  and  state  what 
principle  is  used  in  each  case : 

9.  3  ax -\- 7  ax  —  4:  ax.  15.  (24  a;  +  9  ?/) -r- 3. 

10.  7by-\-4.by-9by.  16.  (6a-46)--2, 

11.  Sy  —  2y-\-y.  17.  (3xy  +  2bx)^x. 

12.  c(4a  +  2  6).  18.  (7  abc  +  2  abd) -r- ab. 

13.  5(16a-3  6).  19.  2a(7 x  +  3y -4:). 

14.  a(ll-&).  20.  3x(;2a-5b  +  G). 

What  principles  are  used  in  the  following  operations  ? 

21.  3.5a6  =  15a6.  23.   3^  +  15^  =  18^. 

22.  16xy^x  =  16y.  24.    78 /i  -  41 /i  =  37 /i. 


7. 

^  +  ^and^  +  ^ 
c               c      c 

8. 

"-''and"     *. 

G                     C        C 

REVIEW   QUESTIONS  17 

29.  Importance  of  the  Principles.  The  five  principles  studied 
in  this  chapter,  together  with  others  which  will  be  introduced 
when  needed,  will  be  found  of  increasing  importance  as  we 
proceed.  Your  success  in  the  further  study  of  algebra  will 
depend  in  no  small  degree  upon  the  clearness  with  which  you 
understand  their  real  significance.  The  most  effective  way  to 
master  them  is  by  means  of  simple  numerical  examples  such 
as  were  used  in  introducing  each  one.  Make  a  list  of  these 
principles  in  abbreviated  form  for  yourself  and  note  how 
frequently  your  own  errors  and  those  of  your  classmates  are 
due  to  direct  violations  of  one  or  more  of  them. 

REVIEW  QUESTIONS 

1.  How  would  3  •  5  and  7  •  5  be  added  in  arithmetic  ?  Why 
cannot  3  n  and  7  n  be  added  in  the  same  manner?  State  in 
full  the  principle  by  which  3  n  and  7  n  are  added.  In  this 
example  what  number  is  represented  by  n  ?  Test  the  identity 
3  ?i  H-  7  ?i  =  10  n  by  substituting  any  convenient  value  for  n. 

2.  What  kind  of  numbers  may  be  added  by  Principle  I? 
Have  the  numbers  ac  and  he  a  common  factor?  What  is 
it  ?  What  is  the  coefficient  of  this  common  factor  in  each  ? 
What  is  the  sum  of  these  coefficients?  Is  the  equality 
ac -\- he  =  (a -\-  h)e  true  no  matter  what  numbers  are  repre- 
sented by  a,  h,  and  c  ?  When  this  can  be  said  of  an  equality, 
what  is  it  called? 

3.  How  is  5  •  9  subtracted  from  11  •  9  in  arithmetic  ?  In 
what  different  manner  may  this  operation  be  performed  ?  AVhy 
is  it  sometimes  necessary  to  perform  subtraction  in  the  second 
way  ?  In  the  identity  31  a;  —  12  a;  =  19  x,  what  number  is  rep- 
resented by  X  ?  Test  the  equality  by  substituting  any  conven- 
ient number  for  x.     Is  this  equality  true  for  every  value  of  a;  ? 

Principle  I  may  be  conveniently  abbreviated  as  follows : 

{ac-\-bc  =  (a-\-b)c, 
\ac  —  bc  =  {a  —  b)c. 


18  INTRODUCTION  TO   ALGEBRA 

4.  How  is  11  -f  3  multiplied  by  4  in  arithmetic  ?  In  what 
different  way  may  this  operation  be  performed  ?  Why  is  it 
sometimes  necessary  to  multiply  in  the  second  way  ?  State  in 
full  the  principle  by  which  a  +  8  is  multiplied  by  7. 

Principle  II  may  be  abbreviated  thus: 

rc(a4-  b)  =ca  4-c6, 
\c{a  —  b)  =  ca  —  cb. 

Notice  that  the  identities  in  Principle  II  are  the  same  as 
those  in  I  read  in  reverse  order. 

5.  How  is  12  + 18  divided  by  6  in  arithmetic  ?  In  what 
different  way  may  this  division  be  performed?  Why  is  it 
sometimes  necessary  to  perform  division  in  the  second  way  ? 
State  in  full  the  principle  used  in  performing  the  operation 
(6  a;  +  9  2/)  -^-  3.     How  do  you  divide  (6  a;  —  9  2/)  by  3  ? 

Principle  III  may  be  abbreviated  thus : 

{a-\-b_a,b     a  —  b  __a     b 
I     k     "  k     k'       k     ~k~k' 

6.  How  is  the  product  2-3.5  multiplied  by  4  in  arithmetic? 
In  what  different  way  may  this  multiplication  be  performed  ? 
Why  is  it  ever  performed  in  the  second  way  ?  What  are 
the  factors  of  the  number  ab  ?  How  is  the  product  of  two 
numbers  multiplied  by  another  number  ?  Should  both  factors 
be  multiplied  by  the  number  or  only  one  ?  Is  it  permissible 
to  multiply  either  one  we  choose  ?  Principle  IV  is  abbreviated 
thus :  ^  ^  ^^^f^^  ^  ^/^^-^  xb  =  ax  (kb). 

7.  Divide  2  •  4  •  6  •  20  by  2  without  first  performing  the 
multiplication  indicated  in  2  •  4  •  6  •  20.  Do  this  in  several 
ways  and  show  that  all  the  quotients  obtained  are  equal.  State 
in  full  the  principle  used. 

Principle  V  is  abbreviated  thus : 

(ab)-^k  =  ^xb  =  ax^' 
k  k 

8.  Contrast  Principles  II  and  IV ;  also  III  and  V. 


ALGEBRAIC   SYMBOLS  19 

Historical  Note.  The  oldest  mathematical  manuscript  in  existence  is 
a  papyrus  now  in  the  British  Museum  which  was  written  by  Ahmes,  an 
Egyptian  priest,  possibly  as  early  as  3000  b.c.  This  contains,  besides  work 
in  arithmetic,  some  very  simple  problems  involving  the  beginnings  of 
algebra.  We  have  no  knowledge  that  the  early  Egyptians  advanced 
beyond  these  simplest  rudiments. 

Singularly  enough,  the  Greeks  failed  to  make  many  important  contribu- 
tions to  algebra,  though  they  greatly  advanced  the  sister  science  geome- 
try. Diophantus,  who  lived  in  Alexandria  about  320  a.d.,  is  the  only 
Greek  whose  contribution  to  algebra  is  worthy  of  note.  Much  of  his 
work,  however,  is  not  of  a  kind  which  can  be  included  in  an  elementary 
book  such  as  this. 

These  beginnings  in  Egypt  and  Greece  were  so  insignificant  that  the 
Hindus  may  be  said  to  have  created  algebra.  Most  of  the  material  con- , 
tained  in  begiimers'  books  such  as  the  present  one  was  known  to  the  Hin- 
dus by  the  year  600  a.d.  The  Arabs  learned  algebra  from  the  Hindus 
and  brought  it  to  Europe  in  the  eighth  century.  It  was  not  until  the 
twelfth  century,  however,  that  the  Arabian  books  on  algebra  were  trans- 
lated into  Latin  and  thus  became  accessible  to  Europeans.    ■ 

The  Arabs  used  very  few  symbols.  Number  expressions  were  written 
out  fully  in  words.  Thus,  instead  of  a;^  _^  lo  x  =  30  they  wrote  A  square 
and  ten  of  its  roots  are  equal  to  thirty-nine.  In  this  respect  they  took  a 
step  backward,  for  the  Hindus,  from  whom  the  Arabs  learned  much  of 
their  algebra,  had  made  considerable  use  of  symbols.  As  we  have  seen 
(page  2)  it  was  not  until  the  middle  of  the  seventeenth  century  that  the 
present  symbols  came  into  general  use. 

Letters  were  used  now  and  then  from  the  earliest  time  as  abbreviations 
for  words  representing  numbers.  But  the  systematic  use  of  letters  for 
this  purpose  began  with  the  great  French  algebraist,  Francois  Vieta 
(1540-1607). 

It  must  not  be  supposed,  however,  that  the  use  of  these  symbols  was 
adopted  suddenly.  It  is  a  curious  fact  of  history  that  people  cling  to  the 
old,  even  though  the  new  is  vastly  superior.  It  was  only  after  Ren6  Des- 
cartes (1696-1650)  had  used  these  symbols  in  his  works  that  they  came  to 
be  used  in  texts  on  algebra  (see  also  page  50). 


CHAPTER   II 

EQUATIONS  AND  PROBLEMS 

The  principles  developed  in  the  last  chapter  will  now  be 
used  in  the  solution  of  equations  and  problems. 

SOLUTION  OF  EQUATIONS 

30.  Definitions.  Equalities  in  which  letters  are  used  as  num- 
ber symbols  are  of  two  kinds,  namely : 

(1)  Identities,  such  as  3(5  x-\-6y)=15x-{-lSy,  which  holds 
for  all  values  which  may  be  assigned  to  x  and  y.     See  §  13. 

(2)  Equalities,  such  as  3  oj  =  18,  which  holds  if,  and  only  if, 
x  =  6. 

The  equality  3  a;  =  18  is  said  to  be  satisfied  by  a;  =  G,  because 
this  value  of  x  reduces  both  members  to  the  same  number,  18. 

Ex.  1.    Is  a; -1-  4  =  9  satisfied  bya;  =  4?  bya;  =  5?  bya;  =  6? 
Ex.. 2.    Is  7a;  +  9  =  3a;  +  25  satisfied  by  x=:2?   by  a;  =  3? 
by  a  =  4  ? 

Ex.3.    Is-  +  3  =  6satisfiedbya;  =  4?  bya;  =  8?  byaj  =  12? 

Ex.4.  Is^-±i-i-2a;-l  =  2a;satisfiedbya;  =  2?  byaj  =  4? 
bya;  =  l?        ^ 

31.  An  equality  which  is  satisfied  only  when  certain  partic- 
ular values  are  given  to  one  or  more  of  its  letters  is  called  a 
conditional  equality  with  respect  to  those  letters. 

E.g.  3  x  +  5  =  35  is  an  equality  only  on  the  condition  that  x  =  10. 
a:  +  y  =  10  is  an  equality  for  certain  pairs  of  values  of  x  and  y  like 
1  and  9,  2  and  8,  3  and  7,  5  and  5,  but  certainly  not  for  all  values  of 
X  and  y ;  for  instance,  not  for  a:  =  3  and  y  =  S. 

20 


SOLUTION  OF  EQUATIONS  21 

32.  A  conditional  equality  is  called  an  equation,  and  a  letter 
whose  particular  value  is  sought  to  satisfy  an  equation  is 
called  an  unknown  in  that  equation. 

The  equations  at  present  considered  contain  only  one  un- 
known. 

33.  To  solve  an  equation  in  one  unknown  is  to  find  the  value 
or  values  of  the  unknown  which  satisfy  it.  Such  a  value  of 
the  unknown  is  called  a  root  or  solution  of  the  equation. 

The  following  examples  illustrate  methods  used  in  solving 
equations. 

Ex.  1.    Solve  the  equation  a;  —  5  =  9.  (1) 

Solution.     This  equation  states  tliat  9  is  5  less  than  the  number  x, 
that  is,  if  9  be  increased  by  5,  the  result  is  x. 
Hence,  x  =  14  is  the  solution  of  equation  (1). 
This  result  may  be  obtained  by  adding  5  to  each  member  of  equation  (1), 

thus  r  r  f^  r 

ar  +  5  -  5  =  9  +  5, 
or  x  =  14.  (2) 

Check.     Substitute  a:  =  14  in  equation  (1)  and  get 

14-5  =  9. 
Hence  the  equation  is  satisfied  by  a:  =  14. 

Ex.  2.    Solve  the  equation  a;  +  7  =  12.  (1) 

Solution.     This  equation  states  that  12  is  7  more  than  the  number  x, 
that  is,  if  12  be  diminished  by  7,  the  result  is  x. 
Hence,  x  =  5  is  the  solution  of  equation  (1). 
This  result  may  be  obtained  by  subtracting  7  from  both  members  of  (1), 

thus  ►.  „  ,^  rr 

a;  +  7-7  =  12-7, 
or  x  =  6.  (2) 

Check.     Substitute  a:  =  5  in  (1)  and  get 

5  +  7  =  12. 
Hence  the  equation  is  satisfied  by  x  =  5. 


22  EQUATIONS  AND  PROBLEMS 

Ex.  3.    Solve  the  equation  ^x  =  7.  (1) 

Solution.  This  equation  states  that  one-third  of  the  number  x  is  7, 
that  is,  X  is  three  times  7,  or  21. 

Hence  a;  =  21  is  the  solution  of  equation  (1). 

This  result  may  he  obtained  hy  multiplying  both  members  of  {I)  by  3, 
thus  3.ix  =  3.7, 

or  X  =  21.  (2) 

Check.     Substitute  a;  =  21  in  (1)  and  get 
1 .  21  =  7. 

Hence  the  equation  is  satisfied  hjx  =  21. 

Ex.  4.    Solve  the  equation  5  a?  =  30.  (1) 

Solution.  This  equation  states  that  5  times  the  number  x  is  30, 
that  is,  X  is  one-fifth  of  30,  or  6. 

Hence,  a:  =  6  is  the  solution  of  the  equation. 

This  result  may  be  obtained  by  dividing  both  members  of  equation  (1) 
%  5,  thus  5^_30 

5  "  5' 
or  x  =  Q.  (2) 

Check.     Substitute  a:  =  6  in  equation  (1)  and  get 

5-6  =  30. 
Hence  the  equation  is  satisfied  by  a:  =  6. 

34.  The  above  examples  illustrate  four  ways  of  operating 
upon  both  members  of  an  equation,  so  as  to  produce  in  each 
case  a  new  equation  satisfied  by  the  same  value  of  the  unknown 
as  the  original  equation. 

Each  of  these  operations  changes  the  value  of  both  members, 
but  changes  them  both  alike. 

In  the  next  example,  equations  (2)  and  (3)  show  two  ways 
of  changing  the  form,  but  not  the  value,  of  one  member  alone,  and 
thus  producing  a  new  equation  satisfied  by  the  same  value  of 
the  unknown  as  the  original  equation. 


SOLUTION   OF   EQUATIONS  23 

Ex.  5.    Solve  the  equation 

w-\-2(w-\-5)=58.  (1) 

By  Principle  II,                          m;  +  2  m;  +  10  =  58.  (2) 

By  Principle  I,                                    3  m;  +  10  =  58.  (3) 

Subtracting  10  from  both  members,       'Sw  =  48.  (4) 

Dividing  both  members  by  3,                     w=  16.  (5) 
Check.     Putting  mj  =  16  in  (1),     16  +  2(16  +  5)  =  16  +  2  •  21  =  58. 

By  substitution  verify  that  w  =  16  also  satisfies  equations  (2),  (3), 
and  (4). 

The  operations  involved  in  passing  from  (1)  to  (2)  and  (2) 
to  (3)  are  called  form  changes  which  leave  the  value  of  the  left 
member  unaltered. 

All  the  operations  involved  in  Principles  I  to  V  are  form 
changes  of  this  character.  See  the  list  at  the  end  of  Chapter  I. 
There  are  other  form  changes  which  will  be  considered  as 
need  arises. 

35.  The  members  of  an  equation  may  be  likened  to  the 
scale-pans  of  a  common  balance  in  which  are  placed  objects  of 
uniform  weight,  say  teupenny  nails.  The  scales  balance  only 
when  the  weights  are  the  same  in  both  pans ;  that  is,  when 
the  number  of  nails  is  the  same. 

If  now  the  scales  are  in  balance,  they  will  remain  so  under 
two  kinds  of  changes  in  the  weights  : 

(a)  When  the  number  of  nails  in  the  two  pans  is  increased  or 
diminished  by  the  same  amount ;  corresponding  to  like  changes 
in  value  of  both  members  of  an  equation. 

(p)  When  the  number  in  each  pan  is  left  unaltered  but  the 
nails  are  rearranged  in  groups  or  piles  in  any  manner;  corre- 
sponding to  form  changes  on  either  member  of  an  equation. 

The  equation,  then,  is  like  a  balance,  and  its  members  are  to  be 
operated  upon  only  in  such  ways  as  to  preseroe  the  balance. 


24  EQUATIONS  AND  PROBLEMS 

EXERCISES 

Solve  the  following  equations  and  explain  each  step  involved, 
as  in  the  above  illustrative  examples. 

1.  7.T  =  42.  7.   07-3  =  7.  13.  x-^3x-\-4:X  =  16. 

2.  ll?i  =  77.  8.   2n  +  5  =  lo.  14.  5n-2n  +  3n  =  4.0. 

3.  iy  =  S.  9.  32/  +  8  =  17.  15.  6y  +  Sy-3y  =  S3. 

4.  ■ia;  =  18.         10.   07-4  =  12.  16.  5  07  +  7  +  2  a7  =  14. 

5.  i7i  =  20.         11.  22/-3=:15.  17.  6  7/4-8  2/-5  =  23. 

6.  6w  =  28.         12.  420-5  =  21.  18.  3w-2w-2  =  4. 
The  foregoing  examples  illustrate 

Principle  VI 

36.  Rule.  An  equation  -may  he  changed  into  another 
equation  such  tliat  any  value  of  the  unhnown  which  sat- 
isfies one  also  satisfies  the  other,  by  means  of  any  of  the 
following  operations: 

(1)  Adding  tl%6  same  number  to  both  members ; 

(2)  Subtracting  the  same  number  from  both  members ; 

(3)  Multiplying  both  meinbers  by  any  number  not  zero ; 

(4)  Dividing  both  members  by  any  number  not  zero ; 

(5)  Changing  the  form  of  either  member  in  any  way 
which  leaves  its  value  unaltered. 

The  operations  under  Principle  VI  are  hereafter  referred  to 
in  detail  by  means  of  the  initial  letters,  A  for  addition,  S  for 
subtraction,  J/ for  multipUcatio7i,  D  for  division,  and  F  iov  form 
changes  which  leave  the  value  of  a  member  unaltered. 

Historical  Note.  The  first  Arabic  algebra  known  to  us  dates  from  the 
first  half  of  the  ninth  century.  It  bore  the  name  Al-gebr  walmukabala. 
The  word  al-gebr,  from  which  the  word  algebra  is  derived,  means 
"restoration"  and  refers  to  the  fact  that  the  same  member  may  be 
added  to  or  subtracted  from  each  member  of  an  equation.  The  word 
"walmukabala"  means  the  process  of  simplification  or  form  changes, 
such  as  have  just  been  described.  Thus  it  appears  that  the  Arabs  re- 
garded the  solution  of  equations  as  the  main  business  of  algebra. 


DIRECTIONS  FOR   WRITTEN  WORK  25 

DIRECTIONS  FOR  WRITTEN  WORK 

37.  In  solving  an  equation  the  successive  steps  should  be 
written  as  in  the  following : 

Ex.1.     25(n+l)-f6(4n-3)=50  +  3l7i+2(3-ri)-9.     (1) 

By  F,  using  Principle  II,  we  obtain  from  (1) 

25  n  +  25  +  24  n  -  18  =  50  +  31  n  +  6  -  2  n  -  9.  (2) 

By  F,  using  Principle  I,  we  obtain  from  (2) 

49  n  +  7  =  29  n  +  47.  (3) 

Subtracting  7  and  29  n  f  rona  each  member  of  (3)  and  using  Prin- 
ciple I,  we  have  20  n  =  40.  (4) 
Dividing  each  member  of  (4)  by  20, 

n  =  2.  (5) 

Check.     Substitute  n  =  2  in  equation  (1). 

For  convenience  this  work  can  be  abbreviated  as  follows  : 

25(n  +  1)  +  6(4  n  -  3)  =  50  +  31  n  +  2(3  -  n)  -  9.  (1) 

By  F,  II,     25  n  +  25  +  24  n  -  18  =  50  +  31  n  +  6  -  2  n  -  9.        (2) 
By  F,  I,  49  71  +  7  =  29  n  +  47.  (3) 

By  5 1  7,  29  n,  20  n  =  40.  (4) 

By  Z)  i  20,  n  =  2.  (5) 

5  I  7,  29  n  means  that  7  and  29  n  are  each  to  be  subtracted  from 
both  members  of  the  preceding  equation.  D  \  20  means  that  the 
members  of  the  preceding  equation  are  to  be  divided  by  20. 

Similarly,  in  case  we  wish  to  indicate  tliat  6  is  to  be  added  to  each 
member  of  an  equation,  we  would  write  A  \  6,  and  if  each  member  is 
to  be  multiplied  by  8,  we  would  write  M  \  8.  It  is  important  that  the 
nature  of  each  step  be  recorded  in  some  such  manner. 

Ex.  2.     17n  +  4(2+w)-6  =  o(4  +  n)-54-3n.  (1) 

By  F,  II,  17  n  +  8  +  4  n  -  6  =  20  +  5  n  -  5  +  3  n.  (2) 

By  F,  I,  21  n  +  2  =  15  +  8  n.  (3) 

By  5  I  2,  8  n,  21  n  -  8  n  =  15  -  2.  (4) 

ByF,  I,  13n  =  13.  (5) 

By  D I  13,  n  =  l.  (6) 

Check.     Substitute  n  =  1  in  equation  (1). 


26  EQUATIONS  AND  PROBLEMS 

38.  By  use  of  Principle  VI,  a  term  may  be  transposed  from 
one  member  of  an  equation  to  the  other  by  changing  its  sign. 

E.g.  in  deriving  equation  (4)  from  (3)  in  the  last  solution,  8  n  was 
subtracted  from  both  sides  by  mentally  dropping  it  on  the  right  and 
indicating  its  subtraction  on  the  left.  Likewise  when  2  is  subtracted 
from  both  sides  it  disappears  on  the  left  and  appears  on  the  right  with 
the  opposite  sign.  Each  of  these  indicated  subtractions  might  have 
been  performed  mentally,  thus  writing  equation  (5)  directly  from  (3). 

After  a  little  practice  this  shorter  process  of  transposing  terms  and 
combining  similar  terms  mentally  should  always  be  used. 

39.  An  equation  may  be  translated  into  a  problem.  For  exam- 
ple, the  equation  21ic  +  2  =  8a;H-15  may  be  interpreted  as 
follows :  Find  a  number  such  that  21  times  the  number  plus  2  is 
15  greater  than  8  times  the  number. 

EXERCISES 

Solve  the  following  equations,  putting  the  work  in  a  form 
similar  to  the  above  and  checking  each  result.  Translate  the 
first  five  into  problems. 

1.  13x  +  40-a;  =  88. 

2.  3ic  +  9-f  2a;  +  6  =  18  +  4a;. 

3.  5x-f  3  — a;  =  a;4-18. 

4.  13i/  +  12  +  52/  =  32  +  82/. 

5.  4m  +  6m  +  4:  =  9m  +  6. 

6.  7m  +  18  +  3m  =  12  +  2m  +  38. 

7.  37/+ 4  +  2?/  +  6  =  i/  +  7  +  7/  +  34-30. 

8.  5a.'  +  34-2a;  +  3  =  2iB  +  5  +  3.T  +  3-f-a;. 

9.  2a;  +  4a;4-9-a;  +  6  =  20  +  2a;  +  5H-a;. 

10.  18  +  6m  +  30  +  4m=4m  +  8  +  12  +  3m  +  3-m  +  29. 

11.  2/  +  T2  +  452/  =  106  +  122/. 

12.  42a;  +  56  =  20a;-fl22. 

13.  6a;  +  8  +  a;  +  4  +  5a;  =  7a;  +  32-aj  — 20, 


SOLUTION  OF  PROBLEMS  21 

14.  32x-\-4:-^Tx  =  66  +  3x-[-5x. 

15.  12m  +  3-3?^  =  38  +  2m. 

16.  15m  +  3-2m  +  7  =  3m+60. 

17.  a-f  7  4-3a  =  2a  + 45. 

18.  55  +  30  +  66  =  36  +  150. 

19.  3c  +  18  +  14c  =  6c  +  51. 

20.  17«  +  4  +  3a;  =  7a;  +  30. 

21.  7(m  +  6)  +  107n,  =  42  +  5m  +  24. 

22.  6a;  +  4(4a;  +  2)  +  3(2a;  +  7)  =  85. 

23.  8  +  7(6  +  6n)+2n  =  2(47i  +  5)  +  187i  +  49. 

24.  5(9a;  +  3)+4(3a;  +  2)  =  18ic  +  36. 

25.  7(3a;  +  2)  +  13  =  5(2-rc)  +  43. 

26.  15  +  3(3  +  a;)  +  2(2  +  6a;)=3(3aj  +  4)+28. 

27.  ll(a;  +  5)+3(3a;-l)  =  7(a;  +  2)  +  4a;  +  47. 

SOLUTION  OF  PROBLEMS 

40.  One  great  object  in  the  study  of  algebra  is  to  simplify 
the  solution  of  problems.  This  is  done  by  using  letters  to  repre- 
sent the  unknown  numbers,  by  stating  the  problem  in  the 
form  of  an  equation,  and  by  arranging  the  successive  steps  of 
the  solution  in  an  orderly  manner. 

41.  Illustrative  Problem.  1.  The  shortest  railway  route  from 
Chicago  to  New  York  is  912  miles.  How  long  does  it  take  a 
train  averaging  38  miles  an  hour  to  make  the  journey  ? 

Solution.  Let  /  be  the  number  of  hours  required.  Then  38  t  is  the 
distance  traveled  in  t  hours.  But  912  miles  is  the  given  distance 
traveled. 

Hence  38 1  =  912.  (1) 

By  i)  1 38  <  =  24.  (2) 

Check.     Substitute  /  =  24  in  equation  (1)  and  get 

38.24  =  912. 
Hence  t  =  24:  satisfies  the  conditions  of  the  problem. 


28  EQUATIONS  AND  PROBLEMS 

42.  Illustrative  Problem.  2.  For  how  many  years  must  $  850 
be  invested  at  5  %  simple  interest  in  order  to  yield  $  255  ? 

Solution.     From  arithmetic,  we  have 

principal  x  rate  x  time  =  interest, 
or  prt  =  i.     See  page  3,  Ex.  7. 

Hence,  from  the  conditions  of  this  problem, 

850  X  .05  X  <  =  255,  (1) 

or  42.5^  =  255. 

By  2)  1 42.5  t  =  Q.  (2) 

Check.     Substitute  <  =  6  in  (1)  and  find 
42.5x6  =  255. 
Hence  6  years  satisfies  the  conditions  of  the  problem. 

43.  Illustrative  Problem.  3.  A  boy,  an  apprentice,  and  a 
master  workman  have  the  understanding  that  the  apprentice 
shall  receive  twice  as  much  as  the  boy  and  the  master  work- 
man five  times  as  much  as  the  boy.  How  much  does  each  get, 
if  the  total  amount  received  for  a  piece  of  work  is  $  104. 

Solution.     Let  n  represent  the  number  of  dollars  received  by  the  boy. 
Then,     2n  is  the  number  of  dollars  received  by  the  apprentice, 
and  5  n  is  the  number  of  dollars  received  by  the  master  workman. 

Hence,  n  +  2  n  +  5  7i  and  104  are  number  expressions,  each  repre- 
senting the  total  amount  received. 

Therefore,  n  +  2n  +  5n  =  104.  (1) 

By  Principle  I,  8  n  =  104.  (2) 

Byi)|8,  n  =  13.  (3) 

Hence,  the  amount  received  by  the  boy  is  13  dollars,  by  the  appren- 
tice 26  dollars,  and  by  the  master  workman  65  dollars. 

Check.  By  the  conditions  of  the  problem  the  sum  of  the  amounts 
obtained  should  be  $  104  ;  the  apprentice  should  receive  twice  as  much 
as  the  boy  and  the.  master  workman  five  times  as  much  as  the  boy. 
That  is,  we  should  have 

13  +  26  -H  65  =  104,    26  =  2  •  13  and  65  =  5  •  13. 


SOLUTION   OF  PROBLEMS  29 

PROBLEMS 

Solve  the  following  problems  by  means  of  equations  and 
check  each  result  by  verifying  that  it  satisfies  the  conditions 
of  the  problem. 

1.  Five  times  a  certain  number  equals  80.      What  is  the 
number  ?     Use  n  for  the  number. 

2.  Twelve  times  a  number  equals  132.  What  is  the  number  ? 

3.  A  tank  holds  750  gallons.    How  long  will  it  take  a  pipe 
discharging  15  gallons  per  minute  to  fill  the  tank  ?* 

4.  The  cost  of  paving  a  block  on  a  certain  street  was  S  7  per 
front  foot.   How  long  was  the  block,  if  the  total  cost  was  $4620? 

5.  A  city  lot  sold  for  $  7500.    What  was  the  frontage,  if  the 
selling  price  was  $  225  per  front  foot  ? 

6.  An  encyclopedia  contains  18,000  pages.      How  many 
volumes  are  there,  if  they  average  750  pages  to  the  volume  ? 

7.  For  how  many  years  must   $3500  be  invested  at  6% 
simple  interest  to  yield  $2205? 

8.  At  what  rate  must  $2500  be  invested  for  3  years  in 
order  to  yield  $412.50? 

Suggestion.    By  the  conditions  of  the  problem  2500  x  r  x  3  ='412.50. 

9.  At  what  rate  must  6800  be  invested  for  7  years  in  order 
to  yield  $2380? 

10.  How  many  dollars  must  be  invested  for  5  years  at  44-  % 
simple  interest  to  yield  $  351  ? 

Suggestion.     By  the  conditions  of  the  problem  p  x  .04 J  x  5  =  351. 

11.  How  many  dollars  must  be  invested  for  6  years  at  4J% 
simple  interest  to  yield  $  2422.50? 

12.  A  cut  in  an  embankment  is  500  yards  long  and  4  yards 
deep.     How  wide  is  it  if  18,760  cubic  yards  are  removed? 

Suggestion.  From  arithmetic,  length  x  width  x  height  =  volume 
or  Iwh  =  w. 


30  EQUATIONS   AND  PROBLEMS 

13.  How  deep  is  a  rectangular  cistern  which  holds  500 
cubic  feet  of  water,  if  it  is  6  feet  wide  and  8  feet  long  ? 

14.  How  long  is  a  box  containing  2240  cubic  inches,  if  its 
width  is  14  inches  and  its  depth  10  inches  ? 

15.  If  n  is  a  number,  how  do  you  represent  10  times  that 
number  ? 

16.  If  n  is  a  number,  how  do  you  represent  that  number  plus 
3  times  itself? 

17.  If  n  is  a  number,  how  do  you  represent  5  times  that 
number  plus  3  times  the  number  plus  8  times  the  number  ? 

18.  The  greater  of  two  numbers  is  5  times  the  less,  and 
their  sum  is  180.     What  are  the  numbers  ? 

19.  A  number  increased  by  twice  itself,  4  times  itself,  and 
6  times  itself,  becomes  429.     What  is  the  number  ? 

20.  A  father  is  3  times  as  old  as  his  son,  and  the  sum  of 
their  ages  is  48  years.     How  old  is  each  ? 

21.  In  a  company  there  are  39  persons.  The  number  of 
children  is  twice  the  number  of  grown  people.  How  many 
are  there  of  each  ? 

22.  How  many  dollars  will  amount  to  $620  in  4  years  at 
6  (fo  siihple  interest  ? 

Solution.     From  arithmetic  we  have 

amount  =  principal  +  interest, 
or  a=p  +  i=p+prL 

Hence,  by  the  conditions  of  the  problem, 

620  =p  +  Mx4:X  p. 
By  Principle  II,  620  =  ;>  (1  +  .24) . 

By  A  „:z.-^  =  500. 

23.  Find  what  principal  invested  for  6  years  at  4^  %  simple 
interest  will  amount  to  $1270. 

24.  Find  what  principal  invested  for  12  years  at  5|  %  sim- 
ple interest  will  amount  to  $  4150. 


ALGEBRAIC   KEPRESENTATION   OF   NUMBERS  31 

ALGEBRAIC  REPRESENTATION  OF  NUMBERS 

Skill  in  translating  problems  into  equations  depends  upon 
attention  to  the  following  points : 

(1)  Read  and  understand  clearly  the  statement  of  the  prob- 
lem, as  it  is  given  in  words. 

(2)  Select  the  unknown  number^  and  represent  it  by  a  suitable 
letter,  say  the  initial  letter  of  a  word  which  will  keep  its 
meaning  in  mirid.  If  there  are  more  unknown  numbers  than 
one,  try  to  express  the  others  in  terms  of  the  one  first  selected. 

(3)  Find  two  number  expressions  which,  according  to  the 
problem,  represent  the  same  number,  and  set  them  equal  to 
each  other,  thus  forming  an  equation. 

Special  care  is  needed  in  expressing  the  various  numbers 
involved  in  a  problem  in  terms  of  a  single  unknown  number, 
as  illustrated  in  the  following  exercises. 

1.  If  71  is  a  number,  represent  in  symbols  a  number  7  greater 
than  w;  5  less  than  w;  8  times  as  great  as  n;  one-third  as 
great  as  n. 

2.  Write  in  symbols  n  increased  by  A; ;  n  decreased  by  k ; 
n  multiplied  by  A:;  n  divided  by  k. 

3.  If  the  sum  of  two  numbers  is  10  and  one  of  them  is  a;, 
what  is  the  other  number  ? 

4.  If  two  numbers  differ  by  6  and  the  smaller  is  x,  what  is 
the  other  number  ? 

5.  If  two  numbers  differ  by  6  and  the  greater  is  a:,  what  is 
the  other  number  ? 

6.  If  A  has  m  dollars,  and  B  has  15  dollars  more  than  A, 
how  do  you  represent  B's  money  ?  If  C's  money  is  twice  B's, 
how  do  you  represent  C's  money  ? 

7.  If  n  is  an  integer,  that  is,  a  whole  number,  how  do  you 
represent  the  next  higher  integer  ?  The  second  higher  ?  The 
next  lower?    The  second  lower ?         ,  ,     ^    o 


32  EQUATIONS  AND  PROBLEMS 

8.  What  is  the  value  of  2  ?i,  for  n  =  1,  2, 3, 4, 5,  6,  etc.  ?   li  n 
is  any  integer,  the  number  represented  by  2  n  is  an  even  integer. 

9.  If  2  ?2  is  any  even  integer,  represent  the  next  higher 
even  integer. 

10.  Represent  each  of  four  consecutive  even  integers,  the 
smallest  of  which  is  2  n. 

11.  What  is  the  value  of  2  n  + 1,  for  ti  =  1,  2,  3,  4,  5,  etc.  ? 
If  n  is  any  integer,  the  number  represented  by  2  n  + 1  is  an 
odd  integer. 

12.  If  2?i  +  1  is  any  odd  integer,  represent  the  next  higher 
odd  integer. 

13.  Eepresent  four  consecutive  odd  integers,  the  smallest  of 
which  is  2  n  +  1. 

14.  If  a;  is  a  number,  express  in  terms  of  ic  a  number  5  less  than 
3  times  x  ;  also  a  number  5  times  as  great  as  x  diminished  by  3. 

15.  The  length  of  a  rectangle  is  3  feet  greater  than  its  width. 
If  w  is  the  width,  how  do  you  represent  its  length  ? 

16.  If  w  and  I  are  the  width  and  length  respectively  of  a 
rectangle,  how  do  you  represent  its  perimeter  ?  (The  per- 
imeter of  a  rectangle  means  the  sum  of  the  lengths  of  its 
four  sides.) 

17.  Express  the  perimeter  of  a  rectangle  in  terms  of  its 
width  Wj  if  its  length  is  10  inches  greater  than  its  width. 

18.  Express  the  perimeter  of  a  rectangle  in  terms  of  its 
length  Z,  if  the  length  is  6  inches  greater  than  the  width. 

PROBLEMS 

Check  each  solution  by  finding  whether  the  result  satisfies 
the  conditions  stated  in  the  problem : 

1.  Four  times  a  certain  number  plus  3  times  the  number 
minus  5  times  the  number  equals  48.     What  is  the  number? 

2.  One  number  is  4  times  another,  and  their  difference  is  9. 
What  are  the  numbers  ? 


ALGEBRAIC  REPRESENTATION  OE  NUMBERS     33 

3.  rind  a  number  such  that  when  4  times  the  number  is 
subtracted  from  12  times  the  number,  the  remainder  is  496. 

4.  Thirty-nine  times  a  certain  number,  plus  19  times  the 
number,  minus  56  times  the  number,  plus  22  times  the  num- 
ber, equals  12.     Find  the  number. 

5.  There  are  three  numbers  whose  sum  is  80.  The  second 
is  3  times  the  first,  and  the  third  twice  the  second.  What  are 
the  numbers  ? 

6.  There  are  three  numbers  such  that  the  second  is  11  times 
the  first  and  the  third  is  27  times  the  first.  The  difference 
between  the  second  and  the  third  is  64.     Find  the  numbers. 

7.  There  are  three  numbers  such  that  the  second  is  8  times 
the  first  and  the  third  is  3  times  the  second.  If  the  second 
is  subtracted  from  the  third,  the  remainder  is  48.  Find  the 
numbers. 

8.  The  number  of  representatives  and  senators  together  in 
the  United  States  Congress,  according  to  the  new  apportion- 
ment, is  631.  The  number  of  representatives  is  51  more  than 
4  times  the  number  of  senators.     Find  the  number  of  each. 

9.  The  area  of  Illinois  is  6750  square  miles  more  than  10 
times  that  of  Connecticut.  The  sum  of  their  areas  is  61,640 
square  miles.     Find  the  area  of  each  state. 

10.  The  sum  of  the  horse  powers  of  the  steamships  Olympic 
and  Mauretania  is  116  thousand.  The  Mauretania  has  22 
thousand  horse-power  less  than  twice  that  of  the  Olympic. 
What  is  the  horse-power  of  each  ship  ? 

11.  It  is  twice  as  far  from  Boston  to  Quebec  as  from  Boston  to 
Albany  and  3  times  as  far  from  Boston  to  Jacksonville,  Florida, 
as  from  Boston  to  Quebec.  How  far  is  it  from  Boston  to  each  of 
the  other  three  cities,  the  sum  of  the  distances  being  1818  miles  ? 

12.  Find  three  consecutive  integers  whose  sum  is  144. 

13.  Find  four  consecutive  integers  such  that  twice  the  first 
plus  the  last  equals  48. 

14.  Find  three  consecutive  even  integers  whose  sum  is  54. 


34  EQUATIONS  AND  PROBLEMS 

15.  Find  three  consecutive  even  integers  such  that  3  times 
the  first  is  12  greater  than  the  third. 

16.  Find  two  consecutive  integers  such  that  3  times  the  first 
plus  7  times  the  second  equals  217. 

17.  Find  two  consecutive  integers  such  that  7  times  the 
first  plus  4  times  the  second  equals  664. 

18.  Find  four  consecutive  odd  integers  such  that  7  times 
the  first  equals  5  times  the  last. 

19.  The  perimeter  of  a  square  is  64  inches.  Find  the  length 
of  a  side. 

20.  A  rectangle  is  4  inches  longer  than  it  is  wide.  Find  its 
length  and  width  if  the  perimeter  is  40  inches. 

21.  A  rectangle  is  twice  as  long  as  wide.  Find  its  dimen- 
sions if  the  perimeter  exceeds  the  length  by  60. 

22.  The  length  of  a  rectangle  is  1^  times  as  great  as  its 
width.  Find  its  dimensions  if  the  perimeter  exceeds  the 
width  by  40  inches. 

23.  The  width  of  a  rectangle  is  f  of  its  length  and  the  per- 
imeter exceeds  the  length  by  50  inches.     Find  its  dimensions. 

24.  How  much  must  be  invested  at  6  %  interest  to  amount 
to  $  2650  at  the  end  of  one  year  ? 

25.  How  much  must  be  invested  at  5  %  simple  interest  to 
amount  to  $2025  at  the  end  of  7  years  ? 

26.  At  what  rate  of  interest  per  year  must  $800  be  in- 
vested to  amount  to  $  1000  in  5  years  ? 

27.  Pikes  Peak  is  3282  feet  higher  than  Mt.  ^tna,  and 
Mt.  Everest  is  708  feet  more  than  twice  as  high  as  Pikes  Peak. 
The  sum  of  the  altitudes  of  Mt.  ^tna  and  Mt.  Everest  is 
39,867  feet.     Find  the  altitude  of  each  of  the  three  mountains. 

28.  The  melting  point  of  iron  is  450  degrees  centigrade 
higher  than  5  times  that  of  tin.  Three  times  the  number  of 
degrees  at  which  iron  melts  plus  7  times  the  number  at  which 
tin  melts  equals  6410.     Find  the  melting  point  of  each  metal. 


REVIEW   QUESTIONS  35 

REVIEW  QUESTIONS 

1.  Define  equality  ;  equation ;  identity.  State  in  detail  how 
the  equation  and  the  identity  differ.    Give  an  example  of  each. 

2.  What  value  of  x  satisfies  the  equation  a;  -|-  4  =  9  ? 
What  value  of  x  will  satisfy  the  equation  obtained  by  adding 
7  to  each  member  of  this  equation  ?    by  adding  12  ?    24  ? 

3.  If  4  be  added  to  the  first  member  of  the  equation 
a;  +  4  =  9,  and  6  to  the  second  member,  what  value  of  x  will 
satisfy  the  equation  thus  obtained  ? 

4.  If  the  same  number  is  added  to  each  member  of  an 
equation,  is  the  resulting  equation  satisfied  by  the  same  value 
of  the  unknown  as  the  first  equation  ? 

5.  If  different  numbers  are  added  to  the  members  of  an 
equation,  is  the  resulting  equation  satisfied  by  the  same  value 
of  the  unknown  as  the  first  equation  ? 

6.  If  the  same  number  is  subtracted  from  each  member  of  an 
equation,  is  the  resulting  equation  satisfied  by  the  same  value  of 
the  unknown  as  the  first  equation  ?     Illustrate  by  an  example. 

7.  If  different  numbers  are  subtracted  from  the  members 
of  an  equation,  is  the  resulting  equation  satisfied  by  the  same 
value  of  the  unknown  as  the  first  equation  ?  Illustrate  by  an 
example. 

8.  Ask  and  answer  questions,  similar  to  the  two  preceding, 
about  the  effect  of  multiplying  both  members  of  an  equation  by 
the  same  or  different  numbers.     Illustrate  each  by  examples. 

9.  Ask  and  answer  similar  questions  on  the  effect  of  divid- 
ing the  members  of  an  equation  by  tlie  same  or  different 
numbers.     Illustrate  each  by  an  example. 

10.  Tlie  process  of  solving  an  equation  consists  in  obtaining 
from  it  other  equations  which  are  satisfied  by  the  same  number 
as  the  original  equation. 

Hence  what  operations  may  be  performed  in  solving  an 
equation  ? 

11.  State  Principle  VI  in  full. 


CHAPTER  III 
POSITIVE  AND  NEGATIVE  NUMBERS 

44.  Thus  far  the  numbers  used  have  been  precisely  the  same 
as  in  arithmetic,  though  their  representation  by  means  of 
letters  and  some  of  the  methods  used  in  operating  upon  them 
are  peculiar  to  algebra. 

We  now  proceed  to  the  study  of  a  new  kind  of  number. 

Examples.  What  is  the  highest  temperature  you  have  ever 
seen  recorded  on  the  thermometer  ?  the  lowest  ? 

In  answering  these  questions  you  not  only  give  certain  num- 
bers, but  you  attach  to  each  a  certain  quality.  The  tempera- 
ture is  above  zero  or  below  zero,  that  is,  the  degrees  on  the 
thermometer  are  measured  in  opposite  directions  from  a  start- 
ing point  which  is  labeled  zero. 

45.  It  has  been  found  exceedingly  useful  in  mathematics  to 
extend  the  number  system  of  arithmetic  so  as  to  make  it  apply 
directly  to  cases  like  this.  The  opposite  qualities  involved 
are  designated  by  the  words  positive  and  negative. 

It  is  commonly  agreed  to  call  above  zero  positive  and  below  zero 
negative.  Likewise,  distances  measured  to  the  right  from  a  zero 
point  are  called  positive,  and  those  to  the  left,  negative.     See  §  48. 

46.  Definitions.  The  signs  +  and  "  stand  respectively  for  the 
words  positive  and  negative,  and  numbers  marked  with  these 
signs  are  called  positive  and  negative  numbers  respectively. 

Thus,  5°  above  zero  is  written  +5°,  and  15°  below  zero  is  written  -15°. 
When  no   sign  of   quality  is  written,  the   positive  sign  is 
understood. 

E.g,   +5°  is  usually  written  5°. 


POSITIVE   AND   NEGATIVE   NUMBERS  37 

Positive  and  negative  numbers  are  sometimes  called  signed 
numbers,  because  each  such  number  consists  of  a  numerical 
part,  together  with  a  sign  of  quality  expressed  or  understood. 

The  numerical  part  of  a  signed  number  is  called  its  absolute 
value. 

Thus,  the  absolute  value  of  +3  and  also  of  "3  is  3. 

47.  The  integers  of  arithmetic  may  be  arranged  in  a  series 
beginning  at  zero  and  extending  indefinitely  toward  the  right. 

Thus,  0,  1,  2,  3,  4,  5,  6,  7,  8,  9,  ... 

The  integers  of  algebra  may  be  arranged  in  a  series  be- 
ginning at  zero  and  extending  indefinitely  both  to  the  right 
and  the  left. 

Thus,    ...  -5,  -4,  -3,  -2,  -1,  0,  +1,  +2,  +3,  +4,  +5, .... 

48.  One  of  the  most  extensive  uses  of  signed  numbers  is  for 
marking  the  points  on  a  straight  line.  This  will  also  appear 
later  in  connection  with  the  graph  in  Chapter  VII. 

On  an  unlimited  straight  line  call  some  starting  point  zero, 
and  lay  off  from  this  point  equal  divisions  of  the  line  indefi- 
nitely both  to  the  right  and  to  the  left,  as  shown  in  the  figure. 


6      5      4       3    ""2    ~1       0    "^1 

-\ 1 \ 1 \ — + — I 1 — f H 


In  order  to  describe  the  position  of  any  one  of  these  division 
points,  we  require  not  only  an  integer  of  arithmetic,  to  specify 
how  far  the  given  point  is  from  the  point  marked  zero,  but  also 
a  sign  of  quality  to  indicate  on  which  side  of  this  point  it  is. 

E.g.  +7  marks  the  division  point  7  units  to  the  right  of  zero,  and 
~5  marks  the  point  5  units  to  the  left  of  zero.  Such  a  diagram  is 
called  the  scale  of  signed  numbers. 

Fractions  would  of  course  be  pictured  at  points  between  the  inte- 
gral division  points,  on  the  right  or  the  left  of  the  scale,  according 
as  the  fractions  are  positive  or  negative. 


38  POSITIVE   AND   NEGATIVE   NUMBERS 

ADDITION  OF  SIGNED  NUMBERS 

49.  In  arithmetic  two  numbers  are  added  by  starting  with 
one  and  counting  forivard  the  number  of  units  in  the  other. 

E.g.     To  add  3  to  5  we  start  with  5  and  count  6,  7,  8. 

In  algebra  two  signed  numbers  are  added  in  the  same 
manner  except  that  the  direction,  forward  or  backward,  in 
which  we  count,  is  determined  by  the  sign,  +  or  ",  of  the 
number  which  we  are  adding. 

Thus,  to  add  +5  to  +7,  begin  at  7  to  the  right  of  the  zero  point  in 
the  scale  of  signed  numbers  and  count  5  more  toward  the  right, 
arriving  at  +12.  Thus,  +7  +  +5  =  +12,  which  is  read  :  Positive  7  plus 
positive  5  equals  jwsitive  12. 

To  add  "5  to  "7,  begin  at  7  to  the  left  and  count  5  more  toward 
the  left,  arriving  at  -12.  Thus,  -7  +  -5  =  "12,  read  :  Negative  7  plus 
negative  5  equals  negative  12. 

To  add  ~5  to  +7,  begin  at  7  to  the  right  and  count  5  toward  the 
left,  arriving  at  +2.  Thus,  +7  +  "5  =  +2,  read :  Positive  7  plus  negative 
5  equals  positive  2. 

To  add  +5  to  "7^  begin  at  7  to  the  left  and  count  5  toward  the  right, 
arriving  at  ~2.  Thus,  -7  +  +5  =  "2,  read:  Negative  7  plus  positive  5 
equals  negative  2. 

The  addition  of  positive  and  negative  numbers  is  further 
explained  in  the  following : 

50.  Illustrative  Problems.  1.  If  a  man  gains  S1500  and  then 
loses  %  800,  what  is  the  net  result  ?     Answer,  %  700  gain. 

In  this  case  the  result  is  obtained  by  subtracting  800  from 
1500.  Yet  this  is  not  really  a  problem  in  subtraction  but  in 
addition.  That  is,  we  are  not  asking  for  the  difference  between 
S 1500  gain  and  %  800  loss,  but  for  the  net  result  when,  the  gain 
and  the  loss  are  taken  together,  or  the  sum  of  the  profit  and 
the  loss.  Hence,  we  say  $  1500  gain  +  $  800  loss  =  $  700  gain, 
or  using  positive  and  negative  signs, 

+1500  ^-800  =  +700. 


ADDITION   OF   SIGNED   NUMBERS  39 

2.  The  assets  of  a  commercial  house  are  $  250,000,  and  the 
liabilities  are  S  275,000.  What  is  the  net  financial  status  of 
the  house  ?     Answer j  $  25,000  net  liabilities. 

Thus, 
$250,000  assets  +  $275,000  liabilities  =  $25,000  net  liabilities. 
Or  +250,000  +  -275,000  =  -25,000. 

3.  The  thermometer  rises  18  degrees  and  then  falls  28 
degrees.  What  direct  change  in  temperature  would  produce 
the  same  result  ?     Answer,  10  degrees  fall. 

Thus,  18°  rise  +  28^  fall  =  10°  fall. 

Or  +18  +  -28  =  -10. 

4.  A  man  travels  700  miles  east  and  then  400  miles  west. 
What  direct  journey  would  bring  him  to  the  same  final  desti- 
nation ?     Answer,  300  miles  east. 

Tlius,       700  miles  east  +  400  miles  west  =  300  miles  east. 
Or  +700  +  -400  =  +300. 

EXERCISES 

Perform  the  following  additions 

1.  -3 +  +4.  5.    no +  -10.  9.    -20 +-10. 

2.  7 +  -3.  6.   +8 +  -4.  10.    +12 +  -12. 

3.  +8 +  -16.  7.    -8 +  +17.  11.   -8 +  +8. 

4.  -5  +  ^5.  8.    +5|  +  -3i.  12.    -9 +  -10. 

The  preceding  exercises  illustrate 

Principle  VII 

51.  Rule.  To  add  two  numbers  with  like  signs,  find 
the  sum  of  their  absolute  values,  and  prefix  to  this  their 
common  sign. 

To  add  two  numbers  with  opposite  signs,  find  the  differ- 
ence of  their  absolute  values,  and  prefix  to  this  the  sign  of 
that  one  whose  absolute  value  is  the  greater. 

In  case  their  absolute  values  are  equal,  their  sum  is  zero. 


40  POSITIVE  AND  NEGATIVE  NUMBERS 

52.  Definition.  The  sum  of  two  signed  numbers  thus  ob- 
tained is  called  their  algebraic  sum. 

Hereafter  addition  will  mean  finding  the  algebraic  sum. 

53.  Signed  numbers  find  application  in  any  situation  where 
opposite  qualities  of  the  kind  here  considered  are  present. 
Besides  those  already  mentioned,  other  instances  occur  in  the 
applications  below : 

APPLICATIONS  OF  SIGNED  NUMBERS 

1.  A  balloon  which  exerts  an  upward  pull  of  460  pounds  is 
attached  to  a  car  weighing  175  pounds.  What  is  the  net 
upward  or  downward  pull  ?  Express  this  as  a  problem  in 
addition,  using  positive  and  negative  numbers. 

Solution.  460  lb.  upward  pull  plus  175  lb.  downward  pull  equals 
285  lb.  net  upward  pull.  Using  positive  numbers  to  represent  up- 
ward pull  and  negative  numbers  to  represent  downward  pull,  this 

equation  becomes  .  ^ _  ^ 

^  +460  +  -175  =  +285. 

In  each  of  the  following  translate  the  solution  into  the 
language  of  algebra  by  means  of  signed  numbers  as  in  Ex.  1. 

2.  A  450-pound  weight  is  attached  to  a  balloon  which 
exerts  an  upward  pull  of  600  pounds.  What  is  the  net  up- 
ward or  downward  pull  ? 

3.  A  man's  property  amounts  to  $45,000  and  his  debts  to 
$  52,000.     What  is  his  net  debt  or  property  ? 

4.  The  assets  of  a  bankrupt  firm  amount  to  $  245,000  and  the 
liabilities  to  $325,000.    What  are  the  net  assets  or  liabilities? 

5.  A  man  can  row  a  boat  at  the  rate  of  6  miles  per  hour. 
How  fast  can  he  proceed  against  a  stream  flowing  at  the  rate 
of  2^  miles  per  hour  ?    7  miles  per  hour  ? 

6.  A  steamer  which  can  make  12  miles  per  hour  in  still 
water  is  running  against  a  current  flowing  15  miles  per  hour. 
How  fast  and  in  what  direction  does  the  steamer  move  ? 


AVERAGES   OF  SIGNED   NUMBERS  41 

7.  A  dove  which 'can  fly  40  miles  per  hour  in  calm  weather 
is  flying  against  a  hurricane  blowing  at  the  rate  of  60  miles  per 
hour.     How  fast  and  in  what  direction  is  the  dove  moving  ? 

8.  If  of  two  partners,  one  loses  $  1400  and  the  other  gains 
$  3700,  what  is  the  net  result  to  the  firm  ? 

9.  A  man's  income  is  $2400  and  his  expenses  $1500  per 
year.     What  is  the  net  result  for  the  year  ? 

10.  A  man  loses  $800  and  then  loses  S600  more.  What  is 
the  combined  loss  ?  Indicate  the  result  as  the  sum  of  two  nega- 
tive numbers. 

AVERAGES  OF   SIGNED  NUMBERS 

54.  Half  the  sum  of  two  numbers  is  called  their  average. 
Thus  6  is  the  average  of  4  and  8.  Similarly,  the  average  of 
three  numbers  is  one-third  of  their  sum,  and  in  general  the 
average  of  n  numbers  is  the  sum  of  the  numbers  divided  by  n. 

Find  the  average  of  each  of  the  following  sets : 

1.  10,  12,  14,  16,  18.  3.   7,  9,  11,  13,  15. 

2.  5,  9,  20,  30,  3.  4.   7,  10,  21,  29,  30. 

The  average  gain  or  loss  per  year  for  a  given  number  of  years 
is  the  algebraic  sum  of  the  yearly  gains  and  losses  divided 
by  the  number  of  years. 

Illustrative  Problem.  A  man  lost  $  400  the  first  year,  gained 
$  300  the  second,  and  gained  $  1000  the  third.  What  was  the 
average  loss  or  gain  ? 

Solution.      -400 +  ^300  + ^1000  ^IgOO^^gQQ^ 

3  3 

That  is,  the  average  gain  is  $300. 

PROBLEMS 

1.  Find  the  average  of  $  1800  loss,  $  3100  loss,  $  6800  gain, 
$10,800  loss,  and  $31,700  gain. 

2.  Find  the  average  of  $180  gain,  $360  loss,^  $480  loss, 
$  100  gain,  $  700  gain,  $  400  gain,  $  1300  loss,  $  300  gain,  J  4840 
gain,  and  $  12,000  gain. 


42  POSITIVE   AND   NEGATIVE  NUMBERS 

Mnd  the  average  yearly  temperatures  at  the  following  places, 
the  monthly  averages  having  been  recorded  as  given  below : 

3.  For  New  York  City:  +29°,  +33°,  +39°,  +46°,  +53°,  +63°, 
+67°,  +67°,  +61°,  +52°,  +47°,  +41°. 

4.  For  St.  Vincent,  Minnesota:  "5°,  0°,  +15°,  +35°,  -^55°, 
+60°,  +66°,  +63°,  +55°,  +40°,  +22°,  +5°. 

5.  For  Nerchinsk,  Siberia:  "23°,  "13°,  "10°,  +35°,  +55°,  +70°, 
+70°,  +64°,  +50°,  +30°,  +5°,  -15°. 

SUBTRACTION  OF  SIGNED  NUMBERS 

55.  Definition.  Subtraction  is  the  process  of  finding  the  num- 
ber which  added  to  a  given  number  called  subtrahend  produces 
another  given  number  called  minuend.  The  number  thus 
found  is  called  the  difference  or  remainder. 

E.g.     We  say  8  —  5  =  3,  because  5  +  3  =  8. 

Signed  numbers  may  be  subtracted  by  a  simple  application 
of  this  definition. 

Ex.  1.     +8  -  +5  =  +3,  because  +5  +  +3  =  +8. 

Ex.  2.     -8  -  -5  =  -3,  because  "5  +  "3  =  "8. 

Ex.  3.     -8  -  +5  =  -13,  because  +5  +  -13  =  "8. 

Ex.  4.     +8  -  -5  =  +13,  because  "5  +  +13  =  +8. 

Signed  numbers  may  be  subtracted  by  counting  on  the  num- 
ber scale  in  a  direction  opposite  to  that  indicated  by  the  sign 
of  the  subtrahend. 

Thus  to  subtract  +5  from  +8  begin  with  +8  and  count  5  units  to  the 
left,  and  to  subtract  ~5  from  +8  begin  with  +8  and  count  5  units  to 
the  right. 

56.  A  short  rule  for  subtraction.  Since  +8  —  "5  =  +13,  and 
since  +8  -f  +5  =  +13,  it  follows  that  subtracting  "5  from  +8 
gives  the  same  result  as  adding  +5  to  +8.  Similarly,  "8  —  ~5 
=  -3  and -8 +  +5  =-3. 

Hence,  subtracting  a  negative  number  is  equivalent  to  adding 
a  positive  number  of  the  same  absolute  value. 


1. 

-5  -  -2. 

6. 

2. 

-4-+1. 

7. 

3. 

-5  -  +2. 

8. 

4. 

+3 --5. 

9. 

5. 

+57  -  -32. 

10. 

11. 

-16- +4. 

12. 

+13 --20. 

13. 

-8 --7. 

14. 

19 -+14. 

15. 

-24  -  -19. 

SUBTRACTION   OF   SIGNED   NUMBERS  43 

Since  +8  -  +5  =  +3,  and  since  +8  +  "5  =  +3,  it  follows  that 
subtracting  +5  from  +8  gives  the  same  result  as  adding  ~5  to  +8. 
Similarly,  "8  -  +5  =  "13  and  "8  +  "5  =  -13. 

Hence,  subtracting  a  positive  number  is  equivalent  to  adding 
a  negative  number  of  the  same  absolute  value. 

These  statements  are  illustrated  by  such  facts  as :  Removiiig 
a  debt  is  equivalent  to  adding  property  and  removing  property  is 
equivalent  to  adding  debt. 

Perform  the  following  subtractions  by  changing  the  sign  of 
the  subtrahend  and  adding: 

-32 -+34. 
-52  -  -32. 
-16  -  -12. 
+37  _  +50. 
-23 -+57. 
The  preceding  exercises  illustrate 

Principle  VIII 

57.  Rule.  To  subtract  one  signed  number  from  another 
signed  number,  change  the  sign  of  the  subtrahend  and 
then  add  it  to  tlie  minuend. 

The  change  in  the  sign  of  the  subtrahend  may  be  made  men- 
tally without  rewriting  the  problem.  The  results  are  to  be 
checlted  by  showing  that  the  difference  added  to  the  subtrahend 
equals  the  minuend. 

58.  Subtraction  always  possible.  In  arithmetic  subtraction  is 
possible  only  when  the  subtrahend  is  less  than  or  equal  to  the 
minuend. 

E.g.  In  arithmetic  we  cannot  subtract  5  from  2  since  there  is  no 
positive  number  which  added  to  5  gives  2. 

However,  by  means  of  negative  numbers  we  can  as  easily 
perform  the  subtraction,  2  minus  5,  as  5  minus  2. 
Thus,  2  -  5  =  -3,  since  -3  +  5  =  2. 


1. 

-10 --5. 

2. 

-15  -  +5. 

3. 

+20 --15. 

4. 

+n-+3. 

5. 

-11  -  +5. 

6. 

-17  -  -20. 

13. 

-78  -  -37. 

14. 

+57  _  +84. 

15. 

-48 --31. 

16. 

-39  -  -95. 

17. 

-91-   3. 

18. 

-38  -  +74. 

44  I'oSitiviJ  AND  :fc?t:GATivii:  numbers 

EXERCISES 

Perform  the  following  subtractions  : 

7.  +6 --14. 

8.  +7 --9. 

9.  -11- +6. 

10.  -21 --6. 

11.  +93 -+22. 

12.  +17 --13. 

59.  Double  use  of  the  Signs  +  and  — .  In  §  46,  we  agreed  that 
when  no  sign  of  quality  is  written,  the  sign  +  is  understood. 
Hence  we  may  write: 

+8  +  +5  =  8  +  5.  (1) 

+8-+5  =  8-5.  (2) 

By  Principle  VII,  we  have 

+8  +  -5  =  +8-+5  =  8-5.  (3) 

By  Principle  VIII,  we  have 

+8--5  =  +8  +  +5  =  8  +  5.  •  (4) 

These  examples  show  how  we  may  dispense  with  the  special 
signs  of  quality  +,  or  -,  as  follows : 

1.  Positive  numbers  are  written  without  any  sign  indicating 
quality  except  where  special  emphasis  is  desired,  in  which  case 
the  sign  -f  is  used. 

2.  A  negative  number  when  standing  alone  is  preceded  by 
the  sign  — .     Thus  ~5  is  written  —  5. 

3.  When  a  negative  number  is  combined  with  other  num- 
bers, its  quality  is  indicated  by  the  sign  —  with  a  parenthesis 
inclosing  it. 

Thus  8  +  ~5  is  written  8  +  (  -  5), 

and  8  —  "5  is  written  8  —  (  —  5). 

But  in  such  cases  it  is  customary  to  apply  Principles  VII  and 
VIII  and  write  at  once  8  —  5  instead  of  8  +  (—  5)  and  8  +  5  in- 
stead of  8  —  (—  5). 


SUBTRACTION   OF   SIGNED   NUMBERS  45 

60.  It  is  clear  from  the  examples  in  this  chapter  that  signed 
numbers  are  needed  to  represent  actual  conditions  in  life,  as  in 
case  of  the  thermometer.  While  from  now  on  such  numbers  will 
be  distinguished,  in  accordance  with  universal  custom,  by  the 
signs,  +j  —  >  it  should  be  understood  that  each  of  these  signs 
is  thus  made  to  represent  either  one  of  two  entirely  different 
things,  namely,  an  operation'  or  a  quality. 

However,  after  we  acquire  some  understanding  of  the  matter, 
this  double  use  of  the  signs  seldom  leads  to  any  confusion,  since 
we  can  always  tell  from  the  context  which  use  is  meant.  For 
example,  in  5— 3,  the  sign  —  means  suhtractiouy  while  in  «=  —  3 
it  means  negative. 

But  for  the  sake  of  avoiding  confusion  at  the  outsets,  and  to 
make  clear  that  a  negative  number  is  not  necessarily  a  subtra- 
hend and  that  a  positive  number  is  not  necessarily  an  addend, 
we  have  up  to  this  time  used  the  special  signs  "*",  ~,  which  could 
be  readily  distinguished  from  the  signs  of  addition  and  sub- 
traction, +,  — . 

EXERCISES 

Perform  the  following  indicated  operations : 

1.  9-(-4).  3.    -10  +  6.  5.   6-8-14. 

2.  _10+(-3).       4.   12-4.  6.    -7+8-18. 

7.  Find  the  value  of  a  +  6  if  (1)  a  =  4,  6= -5;  (2)  a=-2, 
6= -7;  (3)  a=-6,  6  =  8;  (4)  a  =  6,  6  =  10. 

8.  Find  the  value  of  a  -  6  if  (1)  a  =  8,  6  =  8 ;  (2)  a  =  -  3 
6  =  -  7  ;  (3)  a=4,  6  =  -  9;  (4)  a  =  -  3,  6  =  6. 

Solve  the  following  equations : 

9.  a;  +  8  =  4.        Suggestion.     Subtract  8  from  each  member. 

10.  a; +  3  =  7.  14.  —  4  +  a;=— 9.  18.  —35  + a;  =  17. 

11.  a;- 9  =  1.  15.  —  5  +  a;  =  4.  19.  17+ a;  =  —  35. 

12.  3  +  a;  =  0.  16.  —  5  +  a;  =  12.  20.  a; -14  =  -18. 

13.  a;  +  13  =  7.  17.  -9  +  a;  =  -18.  21.  a;-25  =  16. 


46  POSITIVE   AND   NEGATIVE   NUMBERS 

MULTIPLICATION   OF   SIGNED  NUMBERS 

61.  The  multiplication  of  signed  numbers  is  illustrated  by 
the  following  problems: 

Illustrative  Problem.  A  balloonist,  just  before  starting,  makes 
the  following  preparations :  (a)  He  adds  9000  cubic  feet  of  gas 
with  a  lifting  power  of  75  pounds  per  thousand  cubic  feet. 
(b)  He  takes  on  8  bags  of  sand,  each  weighing  15  pounds. 
How  does  each  of  these  operations  affect  the  buoyancy  of  the 
balloon  ? 

Solution,  (a)  A  lifting  power  of  75  lb.  is  indicated  by  +  75,  and 
adding  such  a  power  9  times  is  indicated  by  +9.  Hence,  +9  •  (  +  75) 
=  +  675,  or  675  the  total  lifting  power  added. 

(b)  A. weight  of  15  lb.  is  indicated  by  —  15,  and  adding  8  such 
weights  is  indicated  by  +  8.  Since  the  total  weight  added  is  120  lb., 
we  have  +  8  •  (-  15)  =  -  120. 

Illustrative  Problem.  During  the  course  of  his  journey  this 
balloonist  opens  the  valve  and  allows  2000  cubic  feet  of  gas  to 
escape,  and  later  throws  overboard  4  bags  of  sand.  What 
effect  does  each  of  these  operations  produce  on  the  balloon? 

Solution,  (a)  The  gas,  being  a  lifting  power,  is  positive,  but  the 
removal  of  2000  cubic  feet  of  it  is  indicated  by  —2,  and  the  result  is  a 
depression  of  the  balloon  by  150  lb. ;  that  is,  —2  •  (+75)  =  —  150. 

(ft)  The  removal  of  4  weights  is  indicated  by  —  4,  but  the  weights 
themselves  have  the  negative  quality  of  downward  pull.  Hence  to 
remove  4  weights  of  15  lb.  each  is  equivalent  to  increasing  the  buoy- 
ancy of  the  balloon  by  60  lb. ;  that  is,  -  4  •  (-  15)  =  +  60  =  60. 

62.  These  illustrations  of  multiplying  signed  numbers  are 
natural  extensions  of  the  process  of  multiplication  in  arith- 
metic. 

E.g.  Just  as  3.  4  =  4  +  4  +  4  =  12,  so  3.  (-*)=  -4  +  (-4)  +  (-4) 
=  —  12,  and  since  3  •  4  is  the  same  as  +  3  •  (  +  4),  we  write  +3  •  +4 
=  +  12  =  12. 


MULTIPLICATION  OF   SIGNED  NUMBERS  47 

Again,  just  as  we  take  the  multiplicand  additively  when  the 
multiplier  is  a  positive  integer,  so  we  take  it  subtractively  when 
the  multiplier  is  a  negative  integer. 

E.g.  —  3-(-f  4)  means  to  subtract  +4  three  times;  that  is,  to 
subtract  +  12.  But  to  subtract  +  12  is  the  same  as  to  add  — 12. 
Hence,  —  3  •  (+  4)  =  (—  12).  Again,  —  3  •  (  —  4)  means  to  subtract 
—  4  three  times ;  that  is,  to  subtract  —  12.  But  to  subtract  —  12  is 
the  same  as  to  add  +  12.     Hence,  -  3  •  (-  4)  =  +  12*=  12. 

EXERCISES  AND  PROBLEMS 

Explain  the  following  indicated  multiplications  and  find  the 
product  in  each  case : 

1.  _3.(_10).         4.    -7 -(-8).  7.    -5.  (-48). 

2.  10.  (-3).  5.    12.  (-21).  8.    -25.  (-16). 

3.  50.  (-5).  6.    -27.  (-6).  9.    -8-34. 

10.  A  man  gained  $212  each  month  for  5  months,  then  lost 
S 175  per  month  for  3  months.  Express  his  net  gain  or  loss 
as  the  sum  of  two  products. 

11.  A  raft  is  made  of  cork  and  iron.  What  effects  are  pro- 
duced upon  its  floating  qualities  by  the  following  changes? 

(a)  Adding  4  braces,  each  weighing  (under  water)  5  pounds. 

(b)  Removing  3  pieces  of  cork,  each  capable  of  sustaining  3 
pounds,  (c)  Adding  10  pieces  of  cork,  each  capable  of  sus- 
taining 7  pounds. 

The  preceding  exercises  illustrate 

Principle  IX 

63.  Rule.  //  two  numbers  have  the  same  sign,  th^ir 
product  is  positive;  if  they  have  opposite  signs,  their 
product  is  negative. 

In  applying  this  principle  observe  that  the  sign  of  the 
product  is  obtained  quite  independently  of  the  absolute  value 
of  the  two  factors. 

E.g.    |.(-5)=-(Y)=  -3|;   -  12  •  (- 3.5)=  +  42  =  42. 


48  POSITIVE  AND   NEGATIVE   NUMBERS 

64.  The  product  of  several  signed  numbers  is  found  as  illus- 
trated in  the  following : 

-2.5.  (-3)  .  (-4)  .  6=  -10  .  (-3)  .  (-4)  .  6  =  30  .  (-4)  •  6 
=  — 120  .  6  =  —  720.  That  is,  the  first  two  factors  are  multi- 
plied together,  then  this  product  by  the  next  factor,  and  so  on, 
until  all  the  factors  are  multiplied. 

Since  the  product  of  all  positive  factors  is  positive,  the  final 
sign  depends  upon  the  number  of  negative  factors.  If  this 
number  is  even,  the  product  is  positive;  if  it  is  odd,  the 
product  is  negative. 

E.g.  If  there  are  5  negative  factors,  the  product  is  negative;  if 
there  are  6,  it  is  positive. 

In  the  following  exercises  determine  the  sign  of  the  product 
before  finding  its  absolute  value. 

EXERCISES 

1.  _4.3.(-6).(-7). 

2.  -2.  (-3)  -(-5)  .3. 

3.  _5.[-3+(-7)]. 

4.  _5.(_4)  .3.  (-2). 

5.  8.(-9).(-l).(-2). 

6.  -50.  (-20)  .(-30)  .(-40). 

7.  _2.(-3).(-4).m.(-n).i>.(-g). 

DIVISION  OF  SIGNED  NUMBERS 

65.  In  arithmetic  we  test  the  correctness  of  division  by- 
showing  that  the  quotient  multiplied  by  the  divisor  equals  the 
dividend. 

E.g.  27  -^  9  =  3,  because  9  •  3  =  27. 

Hence  division  may  be  defined  as  the  process  of  finding  one 
of  two  factors  when  their  product  and  the  other  factor  are 
given. 


DIVISION   OF   SIGNED   NUMBERS  49 

The  given  product  is  the  dividend,  the  given  factor  the 
divisor,  and  the  factor  to  be  found  is  the  quotient. 

This  definitioti  also  applies  to  the  division  of  signed  num- 
bers. In  dividing  signed  numbers,  however,  we  must  deter- 
mine the  sign  of  the  quotient  as  well  as  its  absolute  value. 

E.g.         -  42  --  (+  6)  =  -  7,  because  -  7  •  (+  6)  =  -  42 ; 
also  -  42  -- (-  G)  =  +  7,  because  +  7  •  (-  6)  =  -  42. 

So  in  every  case  the  test  is : 

Quotient  X  Divisor  =  Dividend. 

In  like  manner  perform  the  following,  and  check  as  above : 

1.  -25-5-5.  4.    -9rs^3. 

2.  —ab-i-a.  5.    Toy -i-  (—15). 

3.  5xy-i-(-x).  6.    —121x^11. 
The  preceding  exercises  illustrate 

Principle  X 
66.   Rule.     T7ie  guoti^rit  of  two  signed  numhers  is  posi- 
tive if  the  dividend  and  divisor  have  like  si£ns,  negative 
if  they  have  opposite  signs. 

EXERCISES 

Perform  the  following  indicated  divisions,  and  check  by 
multiplying  quotient  by  divisor. 


1. 

-28 

7    ■ 

2. 

-42 
-6 

3. 

51 

-17 

4 

-21 

5. 

-75 

5 

6. 

-16 
-1 

7. 

-49 

9. 
10. 


4 -(-9) 

-3" 
-3.8 
-4 


100  •  (-  99) 
"      1    '  '  -25 

8    -^  12     3. (-4). (-6). 8 

"■     -3'         .        •    -1*  -3 

13.   A  man  lost  $300,  $500,  and  $700  during  three  consecu- 
tive months.     Express  his  average  monthly  loss  as  a  quotient. 


50  POSITIVE   AND  NEGATIVE   NUMBERS 

14.  During  five  consecutive  days  the  maximum  temperature 
was  —  5°,  —  8°,  — 10°,  —  4°,  —  6°  respectively.  Find  the  average 
of  these  high  marks.  • 

15.  A  trader  lost  $250  in  each  of  three  months  and  gained 
$75  during  each  of  the  four  succeeding  months.  Find  the 
average  gain  or  loss  for  the  seven  months. 

67.  While  Principles  I-V  were  studied  in  connection  with 
unsigned,  or  arithmetic  numbers  only,  it  is  now  very  impor- 
tant to  note  that  they  all  ap]3ly  to  signed  numbers  as  well. 

In  the  statement  of  these  principles  the  word  number  will 
from  now  on  be  understood  to  refer  either  to  the  ordinary 
numbers  of  arithmetic  or  to  the  signed  numbers,  as  occasion 
may  require.  It  should  also  be  noticed  that  the  numbers  of 
arithmetic  are  used  as  freely  in  algebra  as  in  arithmetic.  It 
is  only  when  we  wish  to  distinguish  them  from  negative 
numbers  that  they  are  called  positive  numbers. 

The  number  system  of  algebra,  so  far  as  we  have  studied 
it  consists  of  the  numbers  of  anthmetic  together  with  the 
negative  numbers. 

Historical  Note.  The  Hindus  appear  to  have  had  quite  clear  notions 
of  a  purely  "negative  number"  as  distinct  from  a  number  to  be  sub- 
tracted. They  recognized  the  difference  between  positive  and  negative 
numbers  by  attaching  to  one  the  idea  of  debt  and  to  the  other  that  of 
assets,  or  by  letting  them  represent  distances  in  opposite  directions.  The 
Arabs,  however,  failed  to  understand  the  negative  numbers  and  did  not 
include  them  in  the  algebra  which  they  brought  to  Europe.  (See  page 
19.)  With  unimportant  exceptions,  until  the  beginning  of  the  seventeenth 
century,  mathematicians  dealt  exclusively  with  positive  numbers. 

The  negative  numbers  were  brought  permanently  into  mathematics  by 
Rene  Descartes.  (See  pages  19  and  114.)  Trying  to  number  all  the 
points  of  a  complete  straight  line,  Descartes  was  compelled  to  start  at  a 
point  and  number  in  both  directions.  Then  it  became  convenient  to 
distinguish  the  numbers  on  the  two  sides  of  this  starting-point  as  positive 
and  negative,  respectively. 

Sir  Isaac  Newton  (1642-1727)  was  the  first  to  let  a  letter  stand  for  any 
number,  negative  as  well  as  positive.     In  such  a  formula  as  a(6  4-  c)  = 


INTERPRETATION   OF  NEGATIVE   NUMBERS  51 

ab  +  «c,  the  predecessors  of  Newton  would  restrict  the  letters  to  represent 
any  positive  numbers,  while  Newton  regarded' the  letters  as  representing 
any  numbers  lohatever  either  positive  or  negative.  This  was  of  very  great 
importance,  since  it  greatly  reduced  the  number  of  formulas  required. 

Negative  numbers  appeared  "absurd"  or  "fictitious"  to  mathemati- 
cians until  they  hit  upon  a  visual  or  graphical  representation  of  them. 
Cajori  in  his  history  of  elementary  mathematics  says:  "Omit  all  illus- 
trations by  lines,  thermometers,  etc.,  and  negative  numbers  will  be  as  ob- 
scure to  modern  students  as  they  were  to  the  early  algebraists."  From 
the  experience  of  the  early  mathematicians  it  would  appear  that  if  the 
pupil  wishes  to  really  understand  positive  and  negative  numbers,  he  must 
study  with  care  applications  such  as  are  given  in  the  first  part  of  this 
chapter. 

INTERPRETATION  AND  USE  OF  NEGATIVE  NUMBERS 

68.  In  solving  a  problem,  a  negative  result  may  have  a 
natural  interpretation  or  it  may  indicate  that  the  conditions 
of  the  problem  are  impossible. 

A  similar  statement  holds  in  reference  to  fractional  answers  in 
arithmetic.  For  example,  if  we  say  there  are  twice  as  many  girls  as 
boys  in  a  schoolroom  and  35  pupils  in  all,  the  number  of  boys  would 
be  35  -f-  3  =  11|,  which  indicates  that  the  conditions  of  the  problem 
are  impossible. 

69.  Illustrative  Problem.  The  crews  on  three  steamers  to- 
gether number  94  men.  The  second  has  40  more  than  the  first, 
and  the  third  20  more  than  the  second.  How  many  men  in 
each  crew  ? 

Solution.         Let  n  =  number  of  men  in  first  crew. 
Then,  n  -1-  40  =  number  of  men  in  second  crew, 

and  »  -[-  40  4-  20  =  number  of  men  in  third  crew. 

Hence,  n  -f  n  -|-  40  +  n  +  40  +  20  =  94, 

and  3  n  -f-  100  =  94. 

3  n  =  -  6. 
n  =  -2. 
Here  the  negative   result  indicates  that  the  conditions  of  the 
problem  are  impossible. 


52  POSITIVE   AND  NEGATIVE   NUMBERS 

70.   Illustrative  Problem.     A  real  estate  agent  gained  $  8400 

on  four  transactions.     On  the  first  he  gained  $6400,  on  the 

second  he  lost  $  2100,  on  the  third  he  gained  $  5000.     Did  he 

lose  or  gain  on  the  fourth  transaction  ? 

Solution.     Since  we  do  not  know  whether  he  gained  or  lost  on  that 

transaction,  we  represent  the  unknown  number  by  n,  which  may  be 

positive  or  negative,  as  will  be  determined  by  the  solution  of   the 

problem. 

Thus  we  have         6400  +  (  -  2100)  +  5000  +  n  =  8400.  (1) 

Hence,  by  VII,  F,  9300  +  n  =  8400.  (2) 

ByS,  n  =  8400-9300.     (3) 

By  Vm,  n  =  -  900.  (4) 

In  this  case  the  negative  result  indicates  that  there  was  a  loss  on 

the  fourth  transaction. 

PROBLEMS 

In  the  following  problems  give  the  solutions  in  full  and  state 
all  principles  used,  together  with  the  interpretation  of  the 
results : 

1.  A  man  gains  $2100  during  one  year.  During  the  first 
three  months  he  loses  $125  per  month,  then  gains  $500  per 
month  during  the  ne:?j:t  five  months.  What  is  the  gain  or  loss 
per  month  during  the  remaining  four  months  ? 

2.  A  man  rowing  against  a  swift  current  goes  9  miles  in  5 
hours.  The  second  hour  he  goes  one  mile  less  than  the  first, 
the  third  two  miles  more  than  the  second,  and  the  fourth  and 
fifth  each  one  mile  more  than  the  third  hour.  How  many 
miles  did  he  go  during  each  of  the  five  hours  ? 

3.  There  are  three  trees  the  sum  of  whose  heights  is  108 
feet.  The  second  is  40  feet  taller  than  the  first,*  and  the 
third  is  30  feet  taller  than  the  second.     How  tall  is  each  tree  ? 

Find  the  average  yearly  temperature  at  each  of  the  following 
places,  the  average  monthly  temperatures  being  as  here  given : 

4.  Port  Conger,  off  the  northwest  coast  of  Greenland  :  —  37°, 
-  43°,  -  32°,  -  15°,  14°,  18°,  35°,  34°,  25°,  4°,  - 17°,  -  30°. 


INTERPRETATION   OF   NEGATIVE   NUMBERS  53 

5.  Franz  Joseph's  Land  :  -  20°,  -  20°,  -  10°,  0°,  15°,  30°, 
35°,  30°,  20°,  10°,  0°,  -  10°. 

6.  North  Central  Siberia:  -  60°,  -  50°,  -  30°,  0°,  15°,  40°, 
40°,  35°,  30°,  0°,  -  30°,  -  50°. 

7.  A  merchant  gained  an  average  of  $2800  per  year  for  5 
years.  The  first  year  he  gained  $  3000,  the  second  $  1500,  the 
third  $  4000,  and  the  fourth  $  2400.  Did  he  gain  or  lose  and 
how  much  during  the  fifth  year  ? 

8.  A  certain  business  shows  an  average  gain  of  $4000 
per  year  for  6  years.  During  the  first  5  years  the  results 
were:  $8000  loss,  $10,000  gain,  $7000  gain,  $3000  gain,  and 
$  12,000  gain.     Find  the  loss  or  gain  during  the  sixth  year. 

9.  A  commercial  house  averaged  $15,000  gain  for  6  years. 
What  was  the  loss  or  gain  the  first  year  if  the  remainmg  years 
show:  $8000  gain,  $24,000  gain,  $2000  loss,  $20,000  gain, 
and  $  50,000  gain,  respectively  ? 

REVIEW  QUESTIONS 

1.  Name  several  pairs  of  opposite  qualities  all  of  which 
are  conveniently  described  by  the  words  positive  and  negor 
tive.  What  symbols  are  nsed  to  replace  these  words  when 
applied  to  numbers  ? 

2.  When  loss  is  added  to  profit,  is  the  profit  increased  or 
decreased?  What  algebraic  symbols  may  be  used  to  distin- 
guish the  numbers  representing  profit  and  loss  ? 

3.  On  the  number  scale  indicate  what  is  meant  by  +  2 ; 
by  —  2.  Indicate  what  is  meant  by  the  sign  +  in  5  +  2 ;  by 
the  sign  —  in  5  —  2 ;  by  the  sign  —  in  a;  =  —  2. 

4.  Why  do  we  call  positive  and  negative  numbers  signed 
numbers  ?     What  is  meant  by  the  absolute  value  of  a  number  ? 

5.  State  Principle  VII  in  full. 

6.  How  is  the  correctness  of  subtraction  tested  in  arith- 
metic ?     Is  the  same  test  applicable  to  subtraction  in  algebra  ? 

7.  Illustrate  the  subtraction  of  positive  and  negative  num- 
bers by  an  example  involving  profit  and  loss. 


i 

54  POSITIVE   AND   NEGATIVE   NUMBERS 

8.  Show  by  counting  on  the  number  scale  that  the  result 
of  subtraction  gives  the  distance  from  subtrahend  to  minuend 
and  that  the  sign  of  the  remainder  shows  the  direction  from 
subtrahend  toward  the  minuend.  For  example,  use  8  —  (—  5) 
and  —  8  —  (-f  5)  to  illustrate  this. 

9.  How  do  negative  numbers  make  subtraction  possible  in 
cases  where  it  is  impossible  in  arithmetic  ? 

10.  What  is  a  convenient  rule  for  subtracting  signed  num- 
bers ?     State  Principle  VIII. 

11.  Write  an  equation  whose  solution  is  a  negative  number. 

12.  Give  an  example  in  which  positive  and  negative  num- 
bers are  multiplied.     State  Principle  IX. 

13.  Define  division.  How  do  we  obtain  the  law  of  signs  in 
division  ?  State  Principle  X.  What  is  the  test  of  the  cor- 
rectness of  division  ? 

14.  Explain  how  one  set  of  signs  +  and  —  can  be  used  to 
indicate  both  quality  and  operation. 

15.  By  means  of  Principles  VII,  VIII,  IX,  and  X,  simplify 

the   expressions,   a +  6,   a  —  b,   a  •  &,  -,  after    substituting   in 

b 

each  various  positive  and  negative  values  of  a  and  b. 

16.  Add  Principles  VII,  VIII,  IX,  and  X  to  the  list  which 
you  made  in  Chapters  I  and  II.  It  is  absolutely  necessary 
that  you  remember  the  rules  stated  in  these  principles.  Any 
short  phrases  that  will  assist  you  in  this  are  of  value.  Fox 
instance  the  following : 

VII.  In  addition,  positive  and  negative  numbers  tend  to  cancel 
each  other.  The  common  sign  or  the  sign  of  the  numerically 
greater  is  the  sign  of  the  result. 

VIII.  In  subtraction,  change  the  sign  of  the  subtrahend  and  add. 

IX.  In  multiplication,  like  signs  give  -\-  and  unlike  signs  — . 

X.  In  division,  like  signs  give  +  and  unlike  signs  give  — . 


DRILL   EXERCISES  56 

DRILL  EXERCISES 

In  solving  the  following  equations  perform  the  required  addi- 
tions or  subtractions  by  the  direct  method  suggested  in  §  38. 

1.  5x+7  —  2x  =  ^x-\-9. 

2.  3  n +  2(71 +  4)  =4  71  +  14. 

3.  S-\-6x-^S(2-{-x)  =  5x-i-2(J. 

4.  4(x  +  3)  +  2(3  a;  + 1)  =  tj(x  +  2)  +  2(x  +  3)  + 19. 

5.  7(x  +  1)  +  3(2  a;  +  3)  =  4(x  +  5)  +  7(a;  +  2)  -  8. 

6.  7(2a:-3)  +  5(4a;-l)  =  3(a;  +  l)  +  2. 

Perform  the  following  indicated  operations : 

7.  _8-(-9)-(+7)+8. 

8.  l6+(-18)-(+2)  +  4. 

9.  12a;+(-7a;)-(-3a;)+2aj. 
10.    34  7i  +  (-30  7i)-(-7ri). 

11.  8.(-3).(-l).  14.    (-12) -[(-2)  ^(-2)]. 

12.  [48^(-6)]^(-l).         15.    (-42).  (-2)  ^(-7). 

13.  3.(-2)(-3)(-l).  16.    l(7ab)^(-a)2'(-S). 

17.  What  is  meant  by  the  average  of  several  numbers  ? 

18.  Find  the  average  of  20,  16,  8,  4,  0,  -  8,  -  12. 

19.  Find  the  average  of  -  8-  32  +  14  +  26  -  40. 

Solve  the  following : 

20.  a;  +  6  =  4.  23.  3a;-8=-16. 

21.  3  a; +  12  =  6.  24.  2(.x- +  6)  =  3(a;  +  5). 

22.  8  +  a;  =  4.  25.  2a;  +  4  =  3a;  +  8. 

26.  If  —  71  represents  a  negative  integer,  how  do  you  repre- 
sent the  next  integer  to  the  right  on  the  number  scale?  the 
next  to  the  left  ? 

27.  If  —  2  71  represents  a  negative  even  integer,  how  do  you 
represent  the  next  even  integer  to  the  right  ?  the  next  to  the 
left  ? 


CHAPTER   IV 
POLYNOMIALS  ' 

71.  We  have  found  that  the  solution  of  problems  leads  us  to 
build  number  expressions  out  of  single  number  symbols. 

E.g.  If  x  is  a  number  representing  my  age  in  years,  then  2(x  —  10) 
is  double  the  number  representing  my  age  10  years  ago. 

Also  2[(a;  —  10)  +  (ar  +  15)]  is  the  number  representing  twice  the 
sura  of  my  ages  10  years  ago  and  15  years  hence. 

Number  expressions  are  now  to  be  studied  more  in  detail. 

72.  Definition.  A  number  expression  composed  of  parts  con- 
nected by  the  signs  +  and  —  is  called  a  polynomial.  Each  of 
the  parts  thus  connected,  together  with  the  sign  preceding  it  is 
called  a  term. 

E.g.  5  a  —  3  a:z/  —  I  rf  +  99  is  a  polynomial  whose  terms  are  5  a, 
—  3  a:?/,  —  ^rt,  and  +  99.     The  sign  +  is  understood  before  5a. 

04..  .  2  4 

Likewise  3a:+-  +  -  is  a  polynomial  whose  terms  are  3  x,  —,  and  -. 

X      y  X  y 

Note.  —  The  yNOYdpohjnomial  is  sometimes  used  in  a  more  restricted 
sense  in  higher  mathematics. 

73.  Definitions.  A  polynomial  of  two  terms  is  called  a  bino- 
mial ;  one  of  three  terms  is  called  a  trinomial.  A  term  taken 
by  itself  is  called  a  monomial. 

E.g.  5a  —  S  xy  is  a  binomial ;  6a  —  Sxy  —  ^rt  is  a  trinomial 
whose  terms  are  the  monomials  5  a,  —  3xy,—  ^ rt. 

According  to  this  definition  a?  +  (6  -f  c)  may  be  called  a  bi- 
nomial though  it  is  equivalent  to  the  trinomial  x-{-b-\-c. 

In  this  case  x  is  called  a  simple  term  and  (h-{-c)  a  compound 
term.  Likewise  we  may  call  3^  +  4a;  —  5 («  +  &)?/  a  trinomial 
having  the  simple  terms  3f,  +4ic,  and  the  compound  term 

-5(a4-&)2/. 

66 


ADDITION  AND   SUBTRACTION   OF   POLYNOMIALS       57 

74.  Definition.  Two  terms  which  have  a  factor  in  common 
are  called  similar  with  respect  to  that  factor. 

E.g.  5  a  and  —  3  a  are  similar  with  respect  to  a;  —  3  xy  and  ~1  x 
are  similar  with  respect  to  z;  5  a  and  —oh  are  similar  with  respect 
to  5;  7  ahc  and  —  |  ahc  are  similar  with  respect  to  abc. 

Similar  terms  may  be  combined  by  Principle  I. 

E.g.  5a-3a=(5-3)a  =  2a;  -3xy-7x= -a:(3y  +  7)  ;  5a-5i= 
5(a-&). 

EXERCISES 

Select  the  common  factor  and  combine  the  similar  terms  in 
each  of  the  following  : 

1.  lx  —  ^X;\-4:X.  8.  3a6  — 2  6c  +  56d 

2.  3  a  — 2  a +  4  a.  9.  7  ax  +  3  5a; -f  12  ca;. 

3.  7a-\-2a  —  5x  +  7x.  10.  5  ax  ■}- S  ax  —  2  ax, 

4.  3  a;  — 2x4- 4  a;  +  2  a;.  H.  2  ar +  2  5r  — 2  cr. 

5.  8r-|-3r  +  2r  — or.  12.  11  rs  — 2  s^  +  4  as. 

6.  9^-3^4-4^-3^.  13.  6ab  +  7  ac-ad. 

7.  ax  4-  6x  4-  ex.  14.  4:xy  —  Syz  —  o  wy. 

ADDITION  AND  SUBTRACTION  OP  POLYNOMIALS 
Addition  of  Polynomials.     In  adding  polynomials  we  use 
Principle  XI 

75.  Rule.  If  two  or  more  terms  are  to  he  added,  they 
may  he  arranged  and  comhined  in  any  desired  order. 

The  truth  of  this  principle  may  be  seen  from  simple  examples : 
Thus,         2  +  3+  5  =  3  +2  -f-  5  =  2  +(3  +  5)  =  (3  +  2)+  5  =  13. 
Also,  8+(-2)+6  =  -2+8  +  6  =  -2+(8+6)=12. 
Make  other  examples  to  illustrate  this  principle. 

Historical  Note.  The  fundamental  character  of  Principle  XI  was  first 
recognized  about  one  hundred  years  ago.  The  principle  as  here  given  com- 
bines in  one  statement  two  laws  of  algebra:  (1)  the  associative  law,  first 
so  called  by  F.  S.  Servois  (1814)  ;  (2)  the  commutative  law,  first  so  called 
by  Sir  William  Hamilton. 


58  POLYNOMIALS 

76.  In  adding  polynomials  the  work  may  be  conveniently 
arranged  by  placing  the  terms  in  columns,  each  column  con- 
sisting of  terms  which  are  similar.  This  is  permissible  by 
Principle  XI. 

Ex.  Add  5x-6y  +  4:Z-[-5at,  -Sx-\-ll7j -16z-9bt, 
and  —  7  2/  +  8  2J. 

Arranging  as  suggested  and  applying  Principles  I  and  VII  we  have 
5x  -    Qy  +    4:z-h  bat 
^Sx-\-lly-l6z-9bt 

-    7y+    82 

2x-    2y  -    4  2  +  (5  a  -  9  6)< 
5  X  and  —  3  a:  are  similar  with  respect  to  their  ci)mmon  factor  x. 
Hence,  by  Principle  I  we  add  the  other  factors  0  and  —  3,  obtaining 
(5-3)x  =  2a:. 

Likewise  we  add  ■{■  bat  and  —  Qht  with  respect  to  the  common 
factor  t,  obtaining  (5  a  —  9&)^  In  the  second  column  the  sum  is 
(—  6  +  11  —  1)y  —  —  2y^  and  in  the  third  column  the  sum  is  (+  4 

Check  by  giving  convenient  values  to  x,  y,  z,  t^  a,  and  b. 
EXERCISES  IN  ADDITION 

1.  Add7  6-3c,+  2d;   _26  +  8c-13d 

2.  Add  6x-Sy  +  4.t-7z;  x-5y-3t',  4ic-42/4-8^; 
Sx  +  2y-St. 

3.  Add  7a  —  4:X-{-12z',  Sa  —  Sx-{-2z]  2a-\-4:X  —  Sz', 
5  a  —  2x  —  4tZ, 

4.  Add5ac  +  86c-4c4-86;  26  +  3  c -2  6c-3ac;  4&-|- 
4c4-&c—  ac;  2  5c  +  4ac-fc;  3&— 4c. 

5.  Add  16  xy  —  13  ccl ;  15  ab  -  2  xy ;  34:  cd  —  3  xy  +  2  ab; 
14:  cd  —  3  xy  —  2  ab. 

6.  Add  34  aa;  4-  4  62/  —  3  z  ;  2by  -\-5z;  3ax  —  7by+5z', 
7  ax -\- 4:  by  —  az. 

7.  Add  3  6  -f-  4  cd  —  2  ae  ;  a6  —  3  cd  +  3  ae ;  3  cd  —  2  a& ; 
4  cd  —  5  tte  +  7  a6. 


ADDITION  AND   SUBTRACTION   OF   POLYNOMIALS       59 

8.  Add    7  ax  — IS  by -\- 5',     9  aa?  +  8  6^  —  4 ;     Shy —12  ax-, 
A  ax +  7  by  —  9. 

9.  Add  oab-3-67  -h5(x-l);  o  •  67  +  3  ab  -  2(x  -  1)  ; 
8(a;  -  1)  -  4  .  67  +  2  a6. 

10.  Add  ll(c  —  9)  +  3(a;  +  ?/)  +  21  wu ;   —  71  wu  —  5{x  +  y) 
-13(c-9). 

11.  Add5(a-f  ?>)-3(c-d);  3(c-(Z)-8(a  +  &);  -2(a  +  6); 
13(c  -  (Z)  -  4(a  +  h). 

12.  Add3+4(c-d)-5(a-6-c);  4(a- 6- c) -f  5(c-d); 
3(a  -  ?>  -  c)  -  9(c  -  d)  4- 12. 

13.  Add   (a-b)-S{c-d)-\-4:{a-b);    5(a- 6)  +  4(c-(^)-h 
7(c  -  d)  -  9(a  -  6). 

14.  Add7(a;-2/)-4(a;4-2/)  +  4-7;    9(x -{- y)  +  3(x  -  y) - 
9-7;  6(a^-2/)  +  2.7-3(aj  +  2/). 

15.  Add  3(i«-5)4-4(c  +  &)+3(a;-.?/);  8(c  + 6)  -  5(.r-2/) 
+  8(a;-5);  7(c  +  ?>)-4(aj- 2/);  3(a;-?/)  +  (a;- 5). 

16.  Add     16(a+b-c)-S(x-y)~^2(a-b)',      2(x-y)- 
3  (a  -  6)  H-  ( a  +  ^>  -  c) ;  7  (a  —  6)  +  4  (a;  -  y)  -  8  (a  +  6  -  c) . 

17.  Add    6(a-6)-5(a;-f-2/)+7(a;-2)-4a6c;    7(a;-z)- 
9(a;  +  ?/)  +  (a-6)  +  2a6c;  \l{a-b)  +  10abc-\-3{x-z)  +  ^x-\-y). 

11.   Subtraction   of   Polynomials.     Since   subtraction   is  per- 
formed by  adding  the  subtrahend,  with  its  sign  changed,  to 
the  minuend.  Principle  XI  permits  us  to  arrange  the  terms  in 
any  desired  order  as  in  addition. 
This  is  illustrated  as  follows : 

From  15  a6  —  17  xy  -f  11  rt  subtract  —  5  aft  +  4  a^  —  5  n^. 
Arranging  as  on  page  58  and  applying  Principles  I  and  VIII : 
15  a&  -  17  x^  +  11  r< 
—  ^ah-\-    4  xy  —    5  nt 
20ab- 21  xy-{-t(nr+5n) 
As  suggested  in  §  57,  it  is  sufficient  to  change  the  signs  of  the  sub- 
trahend mentally,  rather  than  to  rewrite  them  before  adding  to  the 
minuend. 


60  POLYNOMIALS 

EXERCISES  IN  SUBTRACTION 

1.  From  9x-\-3y  —  llz  subtract  —  5x-\-Sy  —  Sz. 

2.  From  12  ab-  3  cd-f- 12  xy  subtract  Sab  -{-2  cd  —  11  xy. 

3.  From  9xc  +  4tad  —  Scz-\-5y  subtract  3y  —  S  ad-^  5  cz. 

4.  From  13  ^  -|-  5  mx  —  5  cv  subtract  2^  —  4  inx  —  3  cv. 

5.  From  3v  —  2w  -\-5  mn  —  4:xz  subtract  —  v  -\-5w  —  3  mn. 

6.  From  31  6  +  4  a;?/  +  16  aa;  —  4  subtract  Sb  —  5xy  —  3ax. 

7.  From  4:  —  3a  —  5xz  —  Svy  —  x  subtract  7  a -\- 2  xz -^  4:  vy. 

8.  From  Sxy  —  3x-\-4,y  subtract  —  2  xy -\-13iv-\-4:X—2y. 

9.  From  2  ab  — 5 -^7  v +  13  abc  subtract  3ab-\-v  +  S  abc. 

10.  From  8  cxa  —  4:yb  —  3yc  subtract  4  cxa  +  2  ?/6  -f-  4  2/c  —  49. 

11.  From  31-45  —  7  xy  subtract  12-45  +  9  xy. 

12.-  From  3  a6c  -  4  +  2(x  -f-  y)  -  3  xy 

subtract  28  +  4  a;?/  —  3(a;  +  y)-\-S  abc. 

13.  From  21  +  9{xy  -z)-\-  3(a  +  b) 

subtract  S(xy  -z)-  8(a  +  &)  + 15. 

14.  From  5  aa;  —  3  &?/  4-  4  aa;  4-  5  6?/  subtract  5  by -\- 3  ax -j- 7  by. 

15.  From  15  •  48  +  8  rt6  -f  49  a;  subtract  7  •  48  -  9  a6  - 14  a;. 

16.  From  19(r  -  5  s)  -f  13(5  x-4:)+7(x-y) 

subtract  17(5  a;  —  4)  —  5(x  —  y)  —  ll(r  —  5  s). 

17.  From  30  +  14(a;  -  5  yz)  -  13(5  y  -  z) 

subtract  32  -{-S{5y  —  z)  —  7(x  —  5  yz). 

18.  From  a(b  +  c)  +  4(m  +  n)  —  16  c 

subtract  9(m  +  n)  +  31c—d{b  +  c). 

19.  From  5(7  .t;- 4)  4- 3(5?/ -3  a;) +  35 

subtract  56  -  9(7  a;  -  4)  +  8(5  2/  -  3  x). 

20.  From  (3a  +  9&-12c)H-3 

subtract  (6  a  - 12  &  - 18  c)  ^  6. 

21.  From  (axy  +  oi/z  —  axz)-r-  a 

subtract  y(x  +  z)  —  2  xz. 


ADDITION  AND   SUBTRACTION  OF   POLYNOMIALS       61 
EXERCISES  IN  ADDITION  AND  SUBTRACTION 

1.  Addox-3y-7r-hSty  -7a;  +  18y-4r-7«,  -20a; 
—  24:y  +  lSr—16t,  and  13  x  +  15y -^11  r +  6t. 

Check  the  sum  by  substituting  x  =  l,  y  =1,  r  =  l,  t  =  l. 

2.  Add  17  a  —  9  6,  3  c  +  14  a,  6  —  3  a,  a  —  17  c,  and  a  — 3  b 
H-4c.     Checkfora  =  l,  6  =  2,  c  =  3. 

3.  Add  2x  +  3y-t,  -6?/  +  8^,  -x  +  y-t,  -4^  +  7a;, 
and  3  y.     Check  for  x  =  2,  y  =  3j  t  =  1. 

4.  Add  17r+4.9-f,  2t  +  Su,  2r-3s  +  4^,  5ic-6t, 
7r  —  3s-\-8u,  and  Sr  —  2t-\-6u.  Check  by  putting  each 
letter  equal  to  1 ;  also  equal  to  2. 

5.  Add  3h  +  2t-\-4:U2i\idh+St-\-3u.  Check  by  putting 
^  =  100,  ^=10,  w  =  l;  I.e.  324 -f  133  =  457. 

6.  Add  4  7i  +  3 <  -j-  w  and  3h-^2t-\-7u.     Check  as  in  5. 

7.  Write  247,  323,  647,  239,  and  41,  as  number  expressions 
like  those  in  Exs.  5  and  6  and  then  add  them. 

8.  Add  647,  391,  276,  and  444  as  in  Ex.  7. 

9.  Add  4:t—  u,  5t  —  Uj  6  t  —  u,  7  t  —  ^i,  and  St—u.  Check 
for  ^  =  10,  w  =  1 ;  also  t  =  l,  u=:l. 

10.  Simplify  :  3  xyz  —  2  xyz  +  5  xyz  —  4  xyz  -|-  xyz  —  xyz. 

11.  Subtract    5a-36  +  6c    from     — 8a  +  76  — He   and 
check. 

12.  From  7  xy  +  ^xz-^^yz  take  17  xy^l^xz  —  20 2/2;. 

13.  From  6  a; -3?/ take  82/- 3  ^. 

14.  From  3/) -4Q'  +  8r  take  7p  — llr  +  llg. 

15.  From  2x  —  3y  take  ^x-{-7 y +  2a —3h. 

16.  From  the  sum  of  ISahc  — 27  xyz +  13  rst  and  —  11  a6c 
+ 16  xyz  —  52  rst  take  67  rst  -  39  a6c. 

17.  To  the  difference  between   the  subtrahend   15  a;  — 18  y 
+  27  2;  and  the  minuend  117  a;  +  97  y  —  81 2;  sfdd  4  a;  —  6  ?/  +  3  2;. 

18.  Add  ll(a;-y)4-15(a-&)   and    -  20(x  -  y) -  37 (a  -  h) 
and  from  the  sum  subtract  135(a;  —  y)—213(a  —  h). 


62  POLYNOMIALS 

NUMBER  EXPRESSIONS  IN  PARENTHESES 

78.  The  sign  +  before  a  parenthesis  means  that  each  term 
within  is  to  be  added  to  what  precedes,  and  the  sign  —  means 
that  each  term  within  is  to  be  subtracted  from  what  precedes. 

By  Principle   VII,  a -\- (-{•  b)  =  a  -{- b  and  a-f-(— 6)  =  a  — 6; 

and  by  Principle  VIII,  a  —  (-\-b)=a  —  b  and  a  —  (—  6)=a  +  6. 

Hence,  we  have 

Principle  XII 

79.  Rule.  A  parenthesis  pj^eceded  by  the  sign  +  may  he 
removed  without  further  change- 

A  parenthesis  preceded  hy  the  sign  —  may  he  removed  hy 
changing  the  sign  of  each  terin  within  it. 

Note  that  in  each  case  the  sign  preceding  the  parenthesis  is 
also  removed  after  the  operation  indicated  by  it  has  been 
carried  out,  and  that  if  no  sign  is  written  before  the  first  term 
in  the  parenthesis,  the  sign  +  is  understood. 

Remove  the  parentheses  and  simplify  the  following : 

Ex.  1.     3  a  +  (a  -  &  +  4)  -(2  a  +  3  6  -  2) 

=  3a  +  a-&+4-2a-36  +  2  =  2a-46  +  6. 

Ex.2.     o(3a:+ ?/)-4(2a:-3y +  2) 

=  15  a:  +  5  y  -  8  a;  +  12  y  -  8  =  7  a;  +  17  y  -  16. 

In  Ex.  2  we  multiply  the  terms  within  the  first  parenthesis 
by  5  and  those  in  the  second  by  4  and  then  remove  the  paren- 
theses by  Principle  XII. 

EXERCISES 

Remove  the  symbols  of  aggregation  and  simplify : 

1.  (3a;-22/)- (4x-f  32/-2). 

2.  x—y—2z—{2>x-\-2y  —  lz). 

3.  3(a  +  6  +  c) -2(a-6  +  c). 

4.  ^(px-y-\-2z)-ll{?^x^y-z). 

5.  5(7^-42/) +9(.'«-2/)-3(2a^  +  32/). 

6.  ^ir-s)  +  (2r-\'S)-{r-2s). 

7.  11  ^  +  (2^-1) -(1-30- 


NUMBER  EXPRESSIONS  IN   PARENTHESES  63 

8.   9(r-s)-3(r  +  s)-|-2(2r-s). 

10.  5x—{S-4.x  +  Ty)+{5x-\-S)-  {5y -\-3x -99). 

11.  -  (3a  +  56-7c)  +  (8a-4c)-(9c-46  4-4a)-91a. 

12.  7-(4-4c  +  2d-2a)4-31o-(4-2a-5d)-(-8c). 

13.  (41  a6  -  21  c  -f  4) -(36  c  +  15-78  a6)  +  (13  c-90  ab-S). 

14.  9by-{4:C-Sby-13)-2c-16-{S4:by-12c-^Sby). 

15.  6 mri 4-(-9m-7  /i4- 14)-  8  ?i  +  (13  mn  -  17  m)  +  34 mn. 

16.  34aa:-(-17aa;4-42)  -\-Sx-  (14a -f  24 a.^- 7). 

17.  19-(  +  2-7a-4  6  4-lla?>)  -(-26  +  8a6-f  4o). 

18.  41  &?/  -  (4  &  -  13  2/  H-  17  6?/)  -  (  -  5  6  -  17  by  -\-lSy). 

19.  39  rs  -  20  s  - 19  r  -  (7  rs  +  8  s  -  19  r)  -  (15  r  -  5  s-  56) 

20.  x-lSx-(2y-3x)-h(2x-4:y)l. 
Suggestion.     First  remove  the  parentheses,  then  the  brace. 

21.  a -\- [a  —  {b -{- c)  —  2  c]. 

22.  a-\-(a-b)-\-{3a-2b)\. 

23.  2a;-3(aj-l)-[ic-2(2a;-l)]. 

24.  a-  la-{-(b-c)  -2(a  +  b-^c)l. 

80.  By  the  converse  of  Principle  XII  terms  may  be  inclosed 
in  a  parenthesis  with  or  without  change  of  sign,  according  as 
the  sign  —  or  +  precedes. 

E.g.  a-{-b  —  c  =  a-{-  (h  ~  c)  and  a  —  b-^c  =  a  —  (h  —  c). 

EXERCISES 

In  each  of  the  following,  place  the  last  three  terms  in  a 
parenthesis. 

1.  a;-?/  +  2-5.  1.  bxy-ox-{-3y  —  2. 

2.  5a  +  36  — c  +  f^.  8.  9aa;4-3  6?/  — 4cd-f2. 

3.  m  —  n-[-p  —  q.  9.  a  —  3b+d  —  hc-\-%. 

4.  5rt-36  4-2c-d  10.  13  -  7a -  36 4- 9c. 

5.  7m-4n  — 3j:>  — (/.  11.  19  a;  — 3  c  +  4  e- 18  («. 

6.  8+46-3c-d  12.  21ax-13bx  +  Mx-^e. 


64  POLYNOMIALS 

PROBLEMS 

See  suggestions  on  solving  problems  on  page  31.  Check  each 
result  by  showing  that  it  satisfies  the  conditions  of  the  problem. 

1.  The  sum  of  two  numbers  is  16.  Seven  times  one  is  8 
less  than  5  times  the  other.     What  are  the  numbers  ? 

2.  From  a  certain  number  a  there  is  subtracted  3  times  the 
remainder  when  8  is  subtracted  from  2  a.  Express  the  result 
in  terms  of  a. 

3.  A  man  invested  a  certain  sum  of  money  at  5  %  simple 
interest.  The  amount  3f  years  later  was  $950.  What  was 
the  investment  ? 

4.  A  man  bought  a  tract  of  coal  land  and  sold  it  a  month 
later  for  $  93,840.  If  his  gain  was  at  the  rate  of  24  %  per 
annum,  what  did  he  pay  for  the  land  ? 

5.  The  melting  point  of  copper  is  250  degrees  (Centigrade) 
lower  than  4  times  that  of  lead.  Ten  times  the  number  of 
degrees  at  which  lead  melts  minus  twice  the  number  at  which 
copper  melts  equals  1152.  What  is  the  melting  point  of  each 
metal  ? 

6.  The  Nile  is  100  miles  more  than  twice  as  long  as  the 
Danube.  Ten  times  the  length  of  the  Danube  minus  4  times  the 
length  of  the  Nile  equals  3400  miles.    How  long  is  each  river  ? 

7.  The  Ganges  River  is  1800  miles  shorter  than  the  Ama- 
zon, and  the  Orinoco  is  300  miles  shorter  than  the  Ganges. 
The  sum  of  their  lengths  is  6900  miles.     How  long  is  each  ? 

8.  Lead  weighs  259  pounds  more  per  cubic  foot  than  cast 
iron,  and  166  pounds  more  than  bronze ;  while  a  cubic  foot  of 
bronze  weighs  807  pounds  less  than  3  cubic  feet  of  iron.  Find 
the  weight  per  cubic  foot  of  each  metal. 

9.  The  world's  gold  production  in  1908  was  29  million 
dollars  less  than  3  times  that  of  1893,  and  the  production  in 
1900  was  59  million  less  than  twice  that  of  1893.  The  pro- 
duction of  1900  and  1908  together  amounted  to  697  million. 
How  much  was  produced  each  year  ? 


PROBLEMS   ON   THE   ARRANGEMENT   OF   DIGITS         65 
PROBLEMS   ON   THE   ARRANGEMENT  AND   VALUE  OF  DIGITS 

81.  If  we  speak  of  the  number  whose  3  digits,  in  order 
from  left  to  right,  are  5,  3,  and  8,  we  mean  538  =  500  +  30+8. 
Likewise,  the  number  whose  three  digits  are  h,  t,  and  u  is  writ- 
ten 100  h-^lOt  +  u. 

Hence,  when  letters  stand  for  the  digits  of  numbers  written  in 
the  decimal  notation,  care  must  be  taken  to  multiply  each  letter 
by  10,  100,  1000,  etc.,  according  to  the  position  it  occupies. 

Illustrative  Problem.  A  number  is  composed  of  two  digits 
whose  sum  is  6.  If  the  order  of  the  digits  is  reversed,  we 
obtain  a  number  which  is  18  greater  than  the  first  number. 
What  is  the  number  ? 

Solution.     Let  x  =  the  digit  in  tens'  place. 

Then,  6  —  a:  =  the  digit  in  units'  place. 

Hence,  the  number  is  10  a:  +  6  —  x.  Reversing  the  order  of  the 
digits,  we  have  as  the  new  number  10(6  —  x)  +  x. 

Hence,  10(6  -  a;)  +  a;  =  18  +  10  x  +  6  -  a;. 

In  each  of  the  examples  1  to  8  below  there  is  a  number  com- 
posed of  two  digits. 

1.  The  digit  in  units'  place  is  2  greater  than  the  digit  in 
tens'  place.  If  4  is  added  to  the  number,  it  is  then  equal  to  5 
times  the  sum  of  the  digits.     What  is  the  number  ? 

2.  The  digit  in  tens'  place  is  3  greater  than  the  digit  in 
units'  place.  The  number  is  one  more  than  8  times  the  sum  of 
the  digits.     What  is  the  number  ? 

3.  The  sum  of  the  digits  is  9.  If  the  order  of  the  digits 
is  reversed,  we  obtain  a  number  which  is  equal  to  12  times 
the  remainder  when~  the  units'  digit  is  taken  from  the  tens' 
digit.     What  is  the  number  ? 

4.  The  sum  of  the  digits  is  12.  If  the  order  of  (^igits  is  re- 
versed, the  number  is  increased  by  18.     Find  the  number. 

5.  The  tens'  digit  is  2  less  than  its  units'  digit.  The  number 
is  1  less  than  5  times  the  sum  of  its  digits.    What  is  the  number  ? 


66  POLYNOMIALS 

6.  The  digit  in  units'  place  is  4  less  than  that  in  tens' 
place.  If  the  order  of  the  digits  is  reversed,  we  obtain  a 
number  which  is  3  less  than  4  times  the  sum  of  the  digits. 
What  is  the  number  ? 

7.  The  digit  in  units'  place  is  2  less  than  twice  the  digit  in 
tens  place.  If  the  order  of  the  digits  is  reversed,  the  number 
is  unchanged.     What  is  the  number  ? 

8.  The  digit  in  tens'  place  is  12  less  than  5  times  the  digit 
in  units'  place.  If  the  order  of  the  digits  is  reversed,  the 
number  is  equal  to  4  times  the  sum  of  the  digits.  What  is 
the  number  ? 

9.  A  number  is  composed  of  three  digits.  The  digit  in 
units'  place  is  3  greater  than  the  digit  in  tens'  place,  which 
in  turn  is  2  greater  than  the  digit  in  hundreds'  place.  The 
number  is  equal  to  96  plus  4  times  the  sum  of  the  digits. 
What  is  the  number  ? 

DRILL  EXERCISES 

1.  Add  3  .^•  4-  4  2/  —  3  2;,  5x  —  2y  —  z,  and  3y  —  5x+7z. 

2.  From  15  a  +  4  6  —  13  6c  subtract  3a  —  Sb+2hc. 

3.  Subtract  7  x—5  y—7  a  from  6x-\-5y-{-3a. 

4.  From  5x  —  4:y  —  9z  subtract  3x  —  Sy-i-2z. 

5.  Add  5a  +  3b  —  2c  and  11  a  —  7  6  +  8  c. 

6.  Add  11  axy  +  13  a;  —  14  ?/,  2  y  —  4  x,  and  3y  -\-x  —  S  axy. 

7.  (5x-3b)-\-{2x  +  b)-{ix-2h-x-^5b). 

8.  Addl9&  +  3c,  26-7c,2c-14&,  andc  +  86. 

9.  -  (a-3b-c)-(2c-a-5b)-i-(a-c  +  b). 

10.  Subtract  2x-\-4:y-\-z  from  13  x  —  3y  — 5  z-{-S. 

11.  5(x-7)+3(14:-x)-^60  =  l-10x. 

12.  13  (1  -  «)  -  6  (2  a;  -  5)  =  80  +  12  x. 

13.  Add  7  a;  — 3?/  — 4,   5x-{-2y-\-5,  and3y  — 8  a;  — 6. 

14.  Add  13  a  +  4  6  -  9  c,  2  c  -  8  6  - 16  a,  and  8  a  -  5  Z>  -  8  c. 

15.  8a;-[2x+3(a;-l)-(2a;-3)]. 

16.  From  17  6  —  4a  —  2c—  19  subtract  8c— 5a  —  86  +  4. 


MULTIPLICATION  OF  POLYNOMIALS  67 

17.  3  _  (3  _  2  4-  6  +  8  -  3)  +  8  -  (9  -  3  +  8). 

18.  3(4:-x)-2{o-6x)=Sx-\-4:. 

19.  12+(2a-3c-46)-(36-c-a-8). 

20.  5  a;  -  (3  a;  -  2  +  2  2/  4-  a;)  + 13  2/  -  (6  -  3  X  +  4). 

21.  Add  2/ -20,  4  2/ +  6,  2  2/ +4  a; -13,  and  2  a; -8  2/ -40. 

22.  Subtract  16  —  a; +  22  —  42/  from  3a;  —  5z— 8y. 

23.  19  +  (2  a;  -  7)  -  (31  -  4  a;  -  8  -  2  .1;)  =  5  a;  +  7. 

24.  16  +  5a;-(8a;  +  9-4a:  +  17)  =8a;-3. 

25.  6a;-3-(4a;  +  8-9a;)-(5a;-2)=:a;+ll. 

MULTIPLICATION  OF   POLYNOMIALS 

The  following  principle  is  useful  in  multiplying  one  mono- 
mial by  another. 

82.  Principle  XIII.  To  obtain  the  product  of  two  or 
more  factors,  these  may  be  arranged  and  multiplied  in 
any  desired  order. 

The  truth  of  this  principle  is  clear  from  examples  such  as : 
2  .  3     5  =  2  •  5  •  3  =  5  .  3  .  2  =  5  .  (3  .  2)  =  2  .  (3  •  5)  =  ;30. 

Historical  Note.  Principle  XIII  like  Principle  XI  states  two  funda- 
mental laws  of  algebra  :  (\)  the  associative  law  of  factors  first  so  called 
by  F.  S.  Servois  ;  (2)  the  commutative  law  of  factors,  first  so  called  by 
Sir  William  Hamilton. 

83.  In  multiplying  algebraic  expressions,  the  same  factor 
frequently  occurs  more  than  once  in  the  same  term. 

Thus  we  may  have  5-5  or  a  •  a.  These  are  written  5^  and 
a^  respectively,  and  read  5  square  and  a  square. 

In  these  expressions  the  2  is  called  an  exponent  and  shows 
that  the  number  above  which  it  is  written  is  to  be  used  tivice 
as  a  factor. 

This  is  a  convenient  way  of  abbreviating  written  expressions. 

E.g.  5  a  •  a  =  5  a2,   5  «  •  3  a  =  (5  .  3)  •  (a  •  a)  =  15  a\ 

7  X    1  x^l  '1  xx  =  7^x2,   ay    ay  =  aayy  =  a^y^. 


68 


POLYNOMIALS 


3-5 

3-8  . 

5-5 
5 

5-8 

8 

84.  The  product  of  two  binomials,  such  as  5  -f  8  and  5  +  3,' 

may  be  obtained  in  two  ways : 

(1)  (5  +  3)(5  +  8)=8.13  =  104. 

(2)  (5  +  3)(5  +  8)  =  5(5  +  8)  +  3(5+8)  =  524-5- 8  +  3 -5  +  3 -8  =  104. 

f^  ft  The    second    method    is   illus- 

trated by  the  accompanying  fig- 
ure in  which  5  +  8  is  the  length 
of  a  rectangle  and  5  +  3  is  its 
width.  The  total  area  is  the 
product  (5  -f  3)  .  (5  -f  8)  and  is 
composed  of  the  four  small 
areas,  5^,  5-8,  3  •  5,  and  3  •  8. 

The  second  method  here  used  for  multiplying  (5  +  3)(5  +  8) 
is  the  only  one  available  when  the  terms  of  the  binomials  can- 
not be  combined. 

Thus 

(x  +  i)(x  +6)  =  X  (x  +  Q)  +  i  (x  -{■  Q)  =  x^  +  6  X  +  4:  X  +  4:  •  Q 
=  x2  +  10  a:  +  24, 
and     (a  +  b)(m  +  n)  =a  (m  +  ji)  -^  b  (m  -\-  ?})  =  am  +  an  +  bm  -{■  bn. 

Hence,  to  multiply  two  binomials,  multiply  each  term  of  one 
by  every  term  of  the  other  and  add  the  products. 

85.  In  a  manner  similar  to  that  just  illustrated  we  may 
multiply  two  trinomials. 

E.g.  The  product  oi  a  ■\- b  +  c  and  wi  4-  n  +  r,  in  which  the  letters 
represent  any  positive  numbers,  may  represent  the  area  of  a  rectangle, 
divided  into  small  rectangles  as  follows : 


am 

an 

ar 

hm 

bn 

br 

cm 

en 

cr 

MULTIPLICATION   OF  POLYNOMIALS  69 

Hence,  the  product  is  : 
(a  +  b  -{•  c)(m  -\-  n  -{-  r)  =  am  +  bm  +  cm  -h  an  +  bn  +  en  +  a7'-\-br  +  cr, 
in  which  eacA  tet'm  of  one  trinomial  is  multiplied  by  every  term  of  the  other 
and  the  products  are  added. 

Evidently  the  same  process  is  apjjlicable  to  the  product  of 
two  such  polynomials  each  containing  any  number  of  terms. 

EXERCISES  AND  PROBLSMS 

Find  the  following  products  : 

1.  {x  +  l){x-^2).       3.    (^4-7)0^  +  4).      5.    (i4-3)(?  +  7). 

2.  {x  +  S){x-\.b).        4.    (a-f8)(a  +  3).      6.    (y-f- 9)(2/  +  2). 

7.  (8  4-l)(8  +  7).  14.    (6s-fl)(s  +  5). 

8.  (s  +  5)(s4-3).  15.    (a;  +  7)(3  a; +4). 

9.  (a-f6)(c  +  d).  16.    (a4-4)(3a  +  l). 

10.  (a;  +  4)(a;  +  3).  17.  (3+a;)(2  +  5a;). 

11.  {x-\-y-\-z){a-\-h-\-c).  18.  (aH- 6)  (3  a  + 7  6). 

12.  (2a;  +  3)(aj  +  2).  19.  {x^y){2x^Sy). 

13.  {o^-x){io-\-x).  20.  (7a;  +  4)(a;  +  8). 

21.  A  rectangle  is  7  feet  longer  than  it  is  wide.  If  its  length 
is  increased  by  3  feet  and  its  width  iucreased  by  2  feet,  its  area 
is  increased  by  60  square  feet.     What  are  its  dimensions  ? 

22.  A  field  is  10  rods  longer  than  it  is  wide.  If  its  length 
is  increased  by  10  rods  and  its  width  increased  by  5  rods,  the 
area  is  increased  by  640  square  rods.  What  are  the  dimen- 
sions of  the  field  ? 

23.  A  farmer  has  a  plan  for  a  granary  which  is  to  be  12  feet 
longer  than  wide.  He  finds  that  if  the  length  is  increased  8 
feet  and  the  width  increased  2  feet,  the  floor  space  will  be 
increased  by  160  square  feet.     What  are  the  dimensions? 

24.  If  the  length  of  a  rectangular  flower  bed  is  increased 
3  feet  and  its  width  increased  1  foot,  its  area  will  be  increased 
by  19  square  feet.  What  are  its  present  dimensions,  if  its 
length  is  4  feet  greater  than  its  width? 


70  POLYNOMIALS 

86.  Polynomials  with  negative  terms.  The  polynomials  mul- 
tiplied in  the  foregoing  exercises  contain  positive  terms  only. 
The  same  process  is  applicable  to  polynomials  containing  nega- 
tive terms,  as  is  seen  in  the  following  examples  : 

Ex.  1.  Find  the  product  of  (7-4)  and  (3-F5).  This 
product,  written  out  term  by  term,  would  give 

[7+(-4)](3  +  5):=7.3  +  7-5+(-4).3+(-4).5 
=  21  +  35  -  12  -  20  =  24. 
Also(7-4)(3  +  5)=3-8  =  24. 
Ex.  2.     Multiply  7-4  and  8-3. 
.       (7-4)(8-3)  =  [7+(-4)][8+(-3)] 

=  7  .  8  +  7  .  (-  3)  +  (-  4)  .  8  +  (-  4)(-  3) 
=  56  _  21  -  32  +  12  =  15. 
Also  (7-4)(8-3)  =  3.5  =  15. 

Similarly,      (x  +  6)(x  -2)=x^-h6x-2x-10  =  x^-Sx-10, 
and  (a:  -  3)(a;  -  5)  =  a;2  -  3  a;  -  5  X  +  15  =  a;2  -  8  a:  +  15. 

EXERCISES 

Perform  the  following  indicated  operations : 

1.  {X'~5){x-3).  9.  (a-b)(c-\-d). 

2.  (a;  — 3)(a;+4).  10.  (a—b){c-d). 

3.  (a-6)(a-l).  •  11.  (x-4:)(x-5). 

4.  (u-\-5)(u  —  3).  12.  (a-h6— c)(m  — n). 

5.  (6  +  2)(6-7).  13.  (a-b)(7a-\-Sby 

6.  (3-6)(4  +  6).  14.  (5-y)(5x  +  3y). 

7.  (3H-a;)(7-3aj).  15.  (2  a -3  6 -fc)(m  + w). 

8.  (n-4:)(3-7i).  16.  (v-t)(7v-5t). 

The  preceding  exercises  illustrate 

Principle  XIV 

87.  Rule.  The  product  of  two  polynoTnials  is  found  by 
multiplying  each  term  of  one  by  every  term  of  the  other, 
and  adding  these  products. 


MULTIPLICATION   OF   POLYNOMIALS  71 

88.  If  there  are  similar  terms  in  either  polynomial,  these 
should  be  added  first,  thus  putting  each  polynomial  in  as  simple 
form  as  possible. 

E.g.    (3  a:  +  2  -  2  a;)(4  a;  +  3  -  3  a:)=:(a;  +  2)(a;  +  3) 

=  a:2+2x+3a:  +  6  =  a:2+5a:  +  6. 

It  should  be  observed  that  Principle  XIV  involves  a  re- 
peated application  of  Principle  II.     Thus 

(a  +  6)  (c  +  d)  =  (a  +  h)c  +  (a  +  h)d  =  ac  +  he  ■\- ad -{-  hd. 

To  simplify  the  process  of  combining  similar  terms  in  the  product, 
it  may  be  found  convenient  to  arrange  the  work  as  in  the  following 
examples : 

Ex.  1.     Multiply  3a;-2by2a;-5. 

Solution.  3  x  -  2 

2x-b 
6a;2-4ar 

-15x  +  10 
6x2 -19  a: +  10 

Ex.2.     Multiply  3a;-2y  +  2by4a;-32/-2. 

Solution.  3a: -2^4- 2 

4a;-3y-2 
V2x^-8xy-\-Sx 

-9xy  +6y^-Qy 

-6x  +  4y-4 


12  a;2  -  17  a;y  +  2  a;  +  6  ^2  _  2  J/  _  4 


EXERCISES  AND  PROBLEMS 

Perform  the  following  indicated  operations : 
In  each  case  simplify  the  expressions  within  the  parentheses 
1  much  as  possible  before  multiplying : 

1.    (a;-7)(3a;+4).     2.  (a? -2) (9 a; +4).     3.   (a-a;)(9a;+4a). 
4.   (5x  +  Sy^4:X-'2y)(6y  +  Sx'-2y-^y), 


72  POLYNOMIALS 

5.  (13a-6-12a)(2  6-3a). 

6.  {xy-5xy  +  4.)(%y-^~ly). 

7.  (ll&  +  3a)(26-35+5). 

8.  (6-4aj  +  3a;)(7a;H-2/-3a;  +  l). 

9.  (a;-2/+3)(5a;-32/  +  5).      10.    (a  - 13  n)(a -n+ 8). 

11.  {x-2  +  y){4.y-Sx). 

12.  (115-a-10&)(6a-3  6-2a). 

13.  (7+2/-^)(2  2/  +  ^-l).        14.    {bx  +  ^y-l){x-2). 

15.  (-7a-l +8a)(5a- 8-3a). 

Solve  the  following  equations  and  check  the  results : 

16.  (aj  +  2)(a;+3)  =  (a;-3)(a;  +  10)  +  10. 

17.  (5.T-4)(6-a?)-97  =  (a;-l)(6-5a;). 

18.  (3  n  -  1)(18  -  7i)  =  (n  +  6) (16  -  3  7i). 

19.  (7-a)(9a-8)  =  31+(36-9a)(a  +  2). 

20.  (4  a  +  4)(a  -  3)  =  (4  a  + 1) (a  +  7)  - 13  a  +  221. 

21.  (7i+6)(3w-4)-14  =  (n+8)(37i-3). 

22.  (8n  +  6)(10-w)+150=(l-7i)(8?i  +  3). 

23.  (a-l)(13-6a)  =  (6a-3)(8-a)-21. 

24.  (7a;-13)(6-a;)-(a;  +  4)(3-7aj)  =  70. 

25.  Find  two  numbers  whose  difference  is  6  and  whose 
product  is  180  greater  than  the  square  of  the  smaller. 

26.  There  are  four  consecutive  even  integers  such  that  the 
product  of  the  first  and  second  is  40  less  than  the  product  of 
the  third  and  fourth.     What  are  the  numbers  ? 

27.  There  are  four  consecutive  integers  such  that  the  prod- 
uct of  the  first  and  third  is  223  less  than  the  product  of  the 
second  and  fourth.     What  are  the  numbers  ? 

28.  Find  four  numbers  such  that  the  second  is  5  greater 
than  the  first,  the  third  5  greater  than  the  second/  and  the 
fourth  5  greater  than  the  third.  The  product  of  the  first  and 
second  is  250  less  than  the  product  of  the  third  and  fourth. 


PROBLEMS   ON   RECTANGLES  AND   TRIANGLES  73 

29.  A  club  makes  an  equal  assessment  on  its  members  each 
year  to  raise  a  certain  fixed  sum.  One  year  each  member  pays 
a  number  of  dollars  equal  to  the  number  of  members  of  the 
club  less  175.  The  following  year,  when  the  club  has  50  more 
members,  each  member  pays  $  5  less  than  the  preceding  year. 
What  was  the  membership  of  the  club  the  first  year  and  how 
much  did  each  pay  ? 

PROBLEMS  ON  RECTANGLES  AND  TRLANGLES 

30.  A  rectangle  is  10  inches  longer  than  wide.  Express  its 
area  in  terms  of  the  width  iv.  If  the  width  is  increased  by  4 
and  the  length  by  6  inches,  express  the  area  in  terms  of  iv. 

31.  A  rectangle  is  8  inches  longer  than  wide.  Express  its 
area  in  terms  of  theVidth  w  after  the  width  isjncreased  4  inches 
and  the  length  decreased  10  inches. 

32.  A  rectangle  is  5  feet  longer  than  it  is  wide.  If  it  were 
3  feet  wider  and  2  feet  shorter,  it  would  contain  15  square  feet 
more.     Find  the  dimensions  of  the  rectangle. 

33.  A  rectangle  is  6  feet  longer  and  4  feet  narrower  than  a 
square  of  equal  area.  Find  the  side  of  the  square  and  the 
sides  of  the  rectangle. 

If  h  is  the  base  of  a  triangle,  h  its  altitude  (height),  and 
a  its  area,  then  area=:  ^(base  x  altitude)-, 

2 

34.  The  base  of  a  triangle  is  2  inches  less  than  its  altitude 
a.     Express  the  area  in  terms  of  a. 

35.  The  altitude  of  a  triangle  is  7  greater  than  its  base  b. 
If  the  altitude  is  decreased  by  8  and  the  base  by  6,  express  its 
area  in  terms  of  b. 

36.  The  altitude  of  a  triangle  is  16  inches  less  than  the 
base.  If  the  altitude  is  increased  by  3  inches  and  the  base 
by  2  inches,  the  area  is  increased  by  52  square  inches.  Find 
the  base  and  altitude  of  the  triangle. 


ba 

62 

a2 

ab 

74  POLYNOMIALS 

SQUARES  OP  BINOMIALS 

89.  Just  as  x^  is  written  instead  of  x  -  x,  so  (a  +  by  is 
written  instead  of  (a  +  6)(a-h6).  The  square  of  a  binomial 
is  found  by  multiplying  the  binomial  by  itself  as  in  §  84. 

E.g.  (a  +  b)(a  +  b)  =  a^  +  ab  +  ab  +  b^  =  a^  +  2  ab  +  b\ 

Hence,  (a  +  by  =  a'-^2ab  +  b\  I 

b This   product  is  illustrated  in  the  accom- 

h       ba        62       panying  figure,  and  is  evidently  a  special  case 

of  the  type  exhibited  in  the  figures,  page  68. 

Translated  into  words  this  identity  is  :   TTie 

square  of  the  sum  of  any  two  numbers  is  equal 

to  the  square  of  the  first  plus  twice  the  product 

of  the  two  numbers  plus  the  square  of  the  second. 

By  formula  I  we  may  square  any  binomial  sum. 

E.g.  Cdx^-2yy  =  {Zxy+2-{^x){2y)+  (2yy=Qx^-^l2xy  +  ^y\ 

EXERCISES 

Find  the  following  products.     Read  the  first  four  at  sight. 

1.  (a +  2)2.  6.    (l+3a)2.  11.    {5x  +  4^y. 

2.  (a;+3)2.  7.    (2a  +  3&)2.  12.   (3a  +  7c)2. 

3.  (m+n)2.  8.    (2  a: +  1)1  13.    (l  +  8a)2. 

4.  (5  +  6)2.  9.    (3  2/ +  2)2.  14.   (x  +  llyy. 

5.  (2  a +  6)2.  10.    (c  +  4  6)2.  15.   (a +  9  6)2. 

90.  Similarly,  we  obtain  the  square  of  the  difference  of  two 
numbers:  {a- bY=  a"" -2ab -^ b\  II 

Translate  this  identity  into  words. 

Ex.     By  use  of  formula  II,  find  the  square  of  a  —  3  6. 

Solution,     (a  -  3  ^>)2  =  a2  _^  2  •  a(-  3  6)  +  (-3  6)2=  a2  -  6  a&  +  9  b\ 

While  these  squares  are  ordinary  products  of  binomials  and  may 
be  found  by  Principle  XIV,  they  are  of  special  importance  and  should 
be  studied  until  they  can  be  given  from  memory  at  any  time. 


SQUARES   OF   BINOMIALS  75 

EXERCISES 

Find  the  following  products.     Read  the  first  five  at  sight: 

1.  (a -3)2.  6.    {1-2  by.  11.   {2x-3yy. 

2.  (6-4)1  7.    (3  c -1)1  12.    {7  c- 3 ay. 

3.  {c-ay.  8.    {2y-3y.  13.    {x-4:yy. 

4.  (x-7)2.  9.    {3  a- 2  by.  14.   (1-4  2)2. 

5.  (3-?0'.  10.    {c-3by.  15.   (a -7  6)2. 
There   are   two  other   special    products   which    should   be 

memorized;  namely,  III  and  IV  below. 
91.   Examples.     Find  the  products, 

(a:  +  2)(x-2),  {x-h5){x-5),  {x-[-9){x-9). 

In  each  of  these  one  factor  is  the  sum  of  two  numbers  and 
the  other  is  the  difference  between  the  same  numbers,  ichile  the 
product  is  the  difference  of  the  squares  of  these  numbers. 
This  is  expressed  by  the  formula 

{x  +  a){x-a)  =  x--a\  III 

By  means  of  this  formula  the  product  of  the  sum  and  dif- 
ference of  any  two  number  expressions  may  be  found. 
E.g.      (3  a  +  2  6)(3  a  -  2  h)  =  (3  ay  -  (2  by  =  9  a^  _  4  J2. 
Verify  this  by  performing  the  multiplication. 

EXERCISES 

Read  the  following  products : 

1.  (a  +  l)(a-l).  9.    (l_7  2/)(l4-7  2/). 

2.  (a  +  3)(a-3).  10.    (tt-4  6)(aH-4  6). 

3.  {k-b){k-^b).  11.  (6a-36)(6a  +  36). 
'4.    {3-x){3-\-x).  12.   (7-9tt)(7-f  9a). 

5.  (2a  +  36)(2a-36).  13.  (2c  +  l)  (2c -1). 

6.  (a  +  2  6)(a-2  6).  14.  (3a  + 6)(3a  -  6). 

7.  (26-l)(26  +  l).  15.  {6k-\-3h){blc-3h). 

8.  {l-\-3x){l-3x).  16.  (9m  +  3n)(9m-3ii). 


76  POLYNOMIALS 

92.   Examples.     Find  the  products  : 

(x  +  2)(x  +  3)  (a;  +  4)(aj-7) 

(a;  -{-5)(x-  2)  (a;  -5)(x-  3) 

EacJi  of  these  products  when  simplified  consists  of  three  termSy 
of  which  the  first  is  a^,  the  last  is  the  product  of  the  second  terms  of 
the  factors,  and  the  coefficient  of  x  in  the  middle  term  is  the  alge- 
braic sum  of  the  second  terms  of  the  factors. 

This  is  expressed  by  the  formula : 

(jr  +  a){x  4-  b)  =jr2+  (a  +  b)x  +  ab.  IV 

Verify  this  by  performing  the  multiplication. 
( 

EXERCISES 

Write  the  following  products.  Also  try  to  read  them  at  sight. 

1.  {x-\-l)(x  +  3).  6.    (a-8)(a  +  10). 

2.  («  +  9)(a;  +  6).  7.    (a  +  7)(a  +  6). 

3.  (2/  +  6)(2/-2).  8.   (a-7)(a  +  6). 

4.  (2/-8)(y/  +  3).  9.    (aft  +  3)(a&  +  7). 

5.  (c_4)(c-2).  10.    (a6-5)(a6-3). 

In  the  formula  (x  -\- d){x -\- h)  =  x^  +{a-\-  b)x  +  ab,  replace 
a  and  b  by  the  following  values  and  simplify  the  results  : 

11.  a  =  5,  6  =  3.  13.    a  =  6,  &=  — 11. 

12.  a  =  8,  6  =—7.  14.    a=  —  5,  &=— 7. 

15.  Find  the  square  of  42  by  writing  it  as  a  binomial,  40+2. 

16.  Square  the  following  numbers  by  writing  each  as  a 
binomial  sum:  51,  53,  93,  91,  102,  202,  301. 

17.  Find  the  square  of  29  by  writing  it  as  a  binomial,  30—1. 

18.  Square  the  following  numbers  by  first  writing  each  as  a 
binomial  diiference:  28,  38,  89,  77,  99,  198,  499,  998,  999. 

19.  Find  the  product  of  41  and  39,  first  indicating  the  prod- 
uct thus,  (40  + 1)  (40  -  1). 


EQUATIONS  AND   PROBLEMS  77 

20.  Find  the  following  products  by  writing  each  pair  of 
factors  as  the  sum  and  difference  of  two  numbers : 

(1)  62.58.  (3)   53.47.  (5)   17.13. 

(2)  27  .  33.  (4)   102  •  98.  (6)   99  •  101. 

EQUATIONS  AND  PROBLEMS 

Solve  the  following  equations,  verifying  except  where 
the  answer  is  given. 

1.  (a4.4)2  4.(a-l)(2a  +  5)  =  (a  +  4)(3a-f 2). 

2.  (a-l)(3a-l)-(a-fl)'-2a2-18. 

3.  (6-a)2  +  (a-3)(2a-5)  =  (3a  +  l)(a-3)-f-84. 

4.  {la-  18)(a  +  4)  -  (a  - 1)^  =  ^{a  +  2)^  -  79. 
6.   (2  6  -  30)(6  -  1)  -  5  6^  =  6  6  -  3(6  +  5)2  +  65. 

6.  (5-c)2  +  (7-c)2-f(9-c)2=(c-l)(3c-58)-93. 

7.  (5c-3)(2  +  c)-4(c-l)2=(c  +  l)'H-54. 

8.  (8  -4c)(5  -c)=^{C'\- ly  +  (c  +  3)(3  c-  8)  4-218. 

9.  (y-l)»  +  4(2/+l)^4-(l-2/)(5y4-6)  =  15y-29. 

10.  a;(a;  +  3)  +  (a;  +  l)(a;+2)=2a;(a?  +  5)  +  2. 

11.  a^  =  (a;-3)(a;  +  6)-12. 

12.  (5-i-5a;)(3-aj)  +  2(a;  +  l)'  +  3(x  +  l)(a;-7)=17(x+l). 

13.  (8  +  3  a;)(4  ^x)-\-{x-  l){x  -  2)  +  2(a;  +  5)^  =  105. 

14.  (5-6)(66  +  5)  +  4(6-3)=^=20-2(6+l)2  +  3  +  16&. 

Answer  Z^^. 

15.  There  is  a  square  field  such  that  if  its  dimensions  are 
increased  by  5  rods,  its  area  is  increased  625  square  rods.  How 
large  is  the  field  ? 

Suggestion  :  If  a  side  of  the  original  field  is  w,  then  its  area  is  w% 
and  the  area  of  the  enlarged  field  is  (w  +  5)^. 

16.  A  rectangle  is  9  feet  longer  than  it  is  wide.  A  square 
whose  side  is  3  feet  longer  than  the  width  of  the  rectangle  is 
equal  to  the  rectangle  in  area.  What  are  the  dimensions  of 
the  rectangle  ? 


78  POLYNOMIALS 

17.  A  boy  has  a  certain  number  of  pennies  which  he  at- 
tempts to  arrange  in  a  solid  square.  With  a  certain  number 
on  each  side  of  the  square  he  has  10  left  over.  Making  each 
side  of  the  square  one  larger,  he  lacks  7  of  completing  it.  How- 
many  pennies  has  he  ? 

18.  A  room  is  7  feet  longer  than  it  is  wide.  A  square  room 
whose  side  is  3  feet  greater  than  the  width  of  the  first  room 
is  equal  to  it  in  area.  What  are  the  dimensions  of  the  first 
room  ? 

19.  Find  two  consecutive  integers  whose  squares  differ 
by  51. 

20.  Find  two  consecutive  integers  whose  squares  differ 
by  97. 

21.  Find  two  consecutive  integers  whose  squares  differ  by  a. 
Show  from  the  form  of  the  equation  obtained  that  a  must  be 
an  odd  integer. 

22.  There  are  four  consecutive  integers  such  that  the  sum 
of  the  squares  of  the  last  two  exceeds  the  sum  of  the  squares 
of  the  first  two  by  20.     What  are  the  numbers  ? 

23.  Two  square  pieces  of  land  require  together  360  rods  of 
fence  ?  If  the  difference  in  the  area  of  the  pieces  is  900  square 
rods,  how  large  is  each  piece  ?    ^{Hint :  a?  —  (90  —  xf  —  900.) 

24.  There  is  a  square  such  that  if  one  side  is  increased  by 
12  feet  and  the  other  side  decreased  by  8  feet  the  resulting 
rectangle  will  have  the  same  area  as  the  square.  Find  the 
side  of  the  square. 

25.  A  regiment  was  drawn  up  in  a  solid  square.  After  50 
men  had  been  removed  the  officer  attempted  to  draw  up  the 
square  by  putting  one  man  less  on  each  side,  when  he  found 
he  had  9  men  left  over.     How  many  men  in  the  regiment  ? 

•  26.  There  is  a  rectangle  whose  length  exceeds  its  width  by 
11  rods.  A  square  whose  side  is  5  rods  greater  than  the  width 
of  the  rectangle  is  equal  to  it  in  area.  What  are  the  dimen- 
sions of  the  rectangle  ? 


REVIEW   QUESTIONS  79 

REVIEW  QUESTIONS 

1.  What  is  a  polynomial  ?  a  term  ?  How  are  poly- 
nomials classified  ?  What  are  similar  terms  ?  By  what  prin- 
ciple are  similar  terms  added  ?  By  what  principle  are  they 
subtracted  ? 

2.  In  adding  or  subtracting  polynomials  how  may  the 
terms  be  arranged  for  convenience  ?  State  the  principle  on 
which  this  is  based  ? 

3.  What  is  the  principle  for  removing  a  parenthesis  when 
preceded  by  the  sign  +  ?  By  the  sign  —  ?  How  may  Prin- 
ciple XII  be  used  for  inclosing  terms  within  a  parenthesis  ? 

4.  In  finding  the  product  of  two  or  more  numbers  how  may 
the  factors  be  arranged  for  convenience  ?  State  the  principle 
on  which  this  is  based. 

5.  Make  a  diagram  to  show  how  to  multiply  (7  -f  4)  by 
(11  +  8)  without  first  uniting  the  terms  of  the  binomials. 
Multiply  (a  -|-  b)  by  (c  -jt-  d)  in  the  same  manner. 

Multiply  (12  —  3)  by  (9  —  7)  in  two  w^ays  and  compare 
results.  State  the  principle  by  which  two  polynomials  are 
multiplied. 

6.  Describe  a  convenient  manner  of  arranging  the  work  in 
multiplying  polynomials.  What  kind  of  terms  in  the  prod- 
uct are  placed  in  the  same  column  ?  Find  the  product  of  7  x— 
3y  -{-1  and  2  x  —  S  y  —  S,  arranging  your  work  this  way. 

7.  State  in  words  what  is  the  square  of  the  binomial  x  +  aj 
of  the  binomial  x  —  a. 

8.  Translate  into  words  the  formula  (x  +  a)(x  —  a)  =.:/?— o?. 

9.  Translate  into  words  the  formula 

{x  -f-  a)(a;  +  6)  =  ar^  +  (a  -j-  6)a;  -f  ah. 
Interpret  this  formula  for  various  positive  and  negative  values 
of  a  and  6. 

10.  Complete  your  list  of  Principles  stated  in  symbols  up  to 
and  including  Principle  XIV. 


80  POLYNOMIALS 

DRILL  EXERCISES 

1.  (oj  -  1)(2  X  -  2)  +  {x  -  5y={3-x){24:  -  3a;)- 7. 

2.  Proni3-4a-5c  +  8a^subtract2a;2-2a-4c  +  8. 

3.  (4:ab-6ac  —  5ad)(b^c-{-d). 

4.  (17  a;  +  3)(a;- 1)  4-  8  =  (2  -  a;)(6  - 17  a;)  + 19. 

5.  5-(a  +  &-c-d-}-8)4-(3  +  a  +  c-d)--5. 

6.  Add6a  +  9,  8a-13,  46ci-8,  and6-54a. 

7.  (a  -  2)(6  a-  4)  +  2(a- 1)^=  (6  -  a)(30  -  8  a)  +  4. 

8.  rrom6(a  +  2)4-3(c  +  4)-2(5-d) 

subtract  2(a  +  2)  -  2(c  +  4)  +  3 (5  -  d). 

9.  Add  12  a^b^c  +  8  aic,  6  ax  -  8  a^ft'c,  and  2  aa;  +  3  a^d^^ 

10.  Add  5  a;?/^  +  3  x^y  -\-  4  a;?/,  2  a^y  —  6  a;?/^  —  3  xy,  and  4  a;y. 

'     11.  Add  6ab^Sc  —  2a,  2c  — 4a6  — 5a,  5c  —  a-\-ab,  and 
3  +  5a-2c-3a6. 

12.  (n  -  4)(6  -  3  n)  _  (6  -  n)'  - 10  =  -  4  n(n  -  4). 

13.  From  35  a6- 8  a;- 9^+13  subtract  16  a6-4«  +  5a;+8. 

14.  (7i  +  2)24-(n-l)2+(n  +  l)'  =  3n(n  +  2)  +  60w  +  130. 
16.  Subtract  5a— 8a;  —  6y  from  13 a; -}- 14 y  — 15 2;  —  4 a. 

16.  From  9y-4a;- 62;-3&  subtract  8  - 9 y - 3 a: - 2  2. 

17.  2  a; +  4 -6(5 a;- 8- 7  a;) +  2 -4 a;  =  6(2 -3  a;) -42. 

18.  -(7  +  4a;-8-2a;)  +  4-2a;  =  6x  +  25. 

19.  (a-l  +  6-c-(f)(4a4-564-3c-2d). 

20.  (4aa;  — 3a?/-f  5a2;  — 8)(a;  +  2/  — 2  +  2). 

21.  (3a-26-f 4c)(2a  +  36-c). 

22.  7-(3a-26-4a)-f6  +  2a-(36-2a-a). 

23.  8 a;  +  (5  y  -  5)-f  (2 y  - 1)-(13  2^  +  8  a; -  17). 


CHAPTER  V 
SIMPLE   FRACTIONS 

93.  Definitions.  In  arithmetic  a  fraction  such  as  f  is  usually 
regarded  as  2  of  the  3  equal  parts  of  a  unit. 

However,  a  fraction  such  as  ^  cannot  be  regarded  in  this 
way,  since  a  unit  cannot  be  divided  into  S^  equal  parts.  ^ 
indicates  that  5  is  to  be  divided  by  3^ ;  i.e.  ^  =  5-^3^. 

In  algebra  any  fraction  is  usually  regarded  as  an  indicated 
division  in  which  the  numerator  is  the  dividend  and  the  de- 
nominator is  the  divisor. 

Thus,  -  is  understood  to  mean  a-i-b. 
b 

The  numerator  and  denominator  are  together  called  the 
terms  of  the  fraction. 

Operations  on  algebraic  fractions  are  performed  in  accordance 
with  the  same  rules  as  in  arithmetic. 

EXBRCISBS 

Supply  the  missing  numerator  in  each  of  the  following: 


1.    1  =  ,.  4.    ?  =  -.  7.        ^ 


24  b     cb  a-1     1-a 


5.     —  = 


4     12  w      3n  a  +  1     (a  +  l)(a-l) 


3.    1  =  _.  6.    i  =  -_.        9.       1 


a     ha  a      -2a  6  +  2      (6  +  3)(6  +  2) 

81 


82  SIMPLE   FRACTIONS 

The  preceding  examples  illustrate 

Principle  XV 

94.  Rule.  Both  terms  of  a  fraction  may  he  multiplied 
or  divided  hy  the  same  number  without  changing  its  value. 

95.  Definition.  The  lowest  common  multiple  (L,  C.  M.)  of  two 
or  more  numbers  is  the  least  number  which  contains  as  factors 
all  the  factors  of  these  numbers. 

E.g.  12  is  the  L.  C.  M.  of  4  and  6.    ahc  is  the  L.  C.  M.  of  ah,  he,  and  ac. 
(a  +  h)  (a  +  1)  is  the  L.  C.  M.  of  (a  +  b)  and  (a  -f  1). 

EXERCISES 

Find  the  L.  C.  M.  of  each  of  the  following: 

1.  3,  5.  5.    a,  hk.  9.    6  +  2,  &  +  3. 

2.  3,  4,  6.  6.    ah,  he,  ac.  10.    1  — a,  1—2 a. 

3.  6,48,24.  7.   3a,  2&,  4c.  11.    2x-\-3,x-4.. 

4.  a,  6.  8.    a-\-l,  a—1.         12.    m  +  3,  m  —  5. 

REDUCTION  TO  A  COMMON  DENOMINATOR 

96.  Examples.  1.  Eeduce  ^  and  J  to  a  common  denominator. 
How  is  the  common  denominator  related  to  the  denominators  2 
and  3? 

2.  Reduce  -  and  -  to  a  common  denominator.     How  is  the 

a  b 

common  denominator  related  to  the  denominators  a  and  b  ? 

2  3  4 

3.  Reduce , , to  a  common  denomi- 

nator.  «  + l'  («+!)(«+ 2)' a  + 2 

Solution.  The  required  denominator  is  the  L.  C.  M.  of  the  given 
denominators ;  that  is,  (a  +  l)(a  +  2). 

Hence,  ^-  =      ^(^  +  ^)        and  -±-  =      ^C^  +  D     . 
a  +  1      (a  +  l)(a  +  2)  a +  2      (a  +  l)(a  +  2) 

already  has  the  required  denominator. 


ADDITION  AND   SUBTRACTION  83 

EXERCISES 

Reduce  each  of  the  following  to  a  common  denominator : 

7      1     1    1 

ab  be  ac 

o     a     b     c 
be  ac  ab 


1. 

1   1 

~a'b 

111 

2. 

a'Vc 

3. 

1      1 
a   —a 

4. 

1.   i, 

aft'  ac 

5. 

a  b_^ 
be  ac 

6. 

2l.  1. 
xy  xz 

f.     m    n     s 
I  «'•    — J  — >  — • 

xy  xz  yz 


10.    ^,  -K-     13.    ^,  ^.     16.        3 


a!  +  l'x  +  2  l-i'A;  — 1  n-4    (»i-l)(m-4) 

11.-2-.,^-.    14.       2-3-,.    17.       1  1 


a-3'a  +  4         '   a;  +  2' a;-2         '   x-l'   {x-l){x-2) 
12.    _^,  _^.       15.    -±^,  -A^.     18.        2  3 


a— 6'  c— d  *   a  — 6'a-t-6  a  —  b'  (a—b)(c—d) 


ADDITION  AND  SUBTRACTION 

97.  Examples.  1.  Add  j  and  |.  How  is  the  numerator  of 
the  sum  found  after  the  fractions  have  been  reduced  to  a 
common  denominator. 

1  2 

2.    Add and 


a  +  1  a  +  3 

Solution.    -L-  = ^L±3 .  _2_  ^       2(a  +  l)      . 

a  +  1      (a  +  l)(a  +  3)    a  +  3      (a+l)(a  +  3) 

Hence, -l-  +  -2_  = ^±1 +        ^(a  +  1) 

'a  +  1      a +  3      (a  +  l)(a  +  3)      (a  +  l)(a  +  3) 

_a  +  3  +  2ra  +  l)_  3a  +  4 

(a  +  l)(a  +  3)         (a  +  l)(a  +  3)" 


84  SIMPLE  FKACTIONS 

EXERCISES 

Perform  the  following  additions : 

1.  i  +  l.  5.    2+^. 
a     0  a      —a 

2.  -1  +  ^-  6.        1       • 


ab     ac  1  —  k     k  —  1 

1     +^-  T.    ^+  2 


ct-f-l      a  — 1  n  — 4     (ri  — l)(n— 4) 

2,3.  «         2       .      3 

4.    — ^4- 


x  +  1      x  +  2  x  +  2      x  —  2 

98.  Examples.  1.  From  ^  subtract  |.  How  is  the  numera- 
tor of  the  difference  found  after  the  fractions  have  been 
reduced  to  a  common  denominator  ? 

3  2 

2.   From subtract 


x-4:  a;  +  3 

Solution : 

_3 2      ^        3(a:+3) 2(a:-4) 

a; -4      a; +  3      (a;-4)(a;  +  3)      (x-4)(a;-3) 

^3(a:  +  3)-2Ca:-4)^         a:  +  17 

(a:_4)(a:-3)  (a;  -  4) (a;  -  3) ' 

EXERCISES 

Perform  the  following  subtractions : 


1. 

1      1 

a      b 

2. 

ab     ac 

3. 

1 

a-Hl     a 

1 

1 

4. 

2 

3 

5. 

2        1 

—  a      a 

6. 

1            1 

1-k     k-1 

7. 

2           3 

iB  +  2     x-2 

8. 

3                   2 

a;  +  laj  +  2  n-4      (n-l)(n-4)  . 


MULTIPLICATION   AND   DIVISION  85 

BXBRCISBS 
Perform  the  following  additions  and  subtractions : 

1.  _^-^.  10.   ^ ^+1. 

a-\-b     a  —  b     a 

2.  -^ --,.  11.    -^  +  -^+1. 

1 —a      a— 1     a 

3.  --.-^.  12.    -1—1  +  1. 
aj  —  l      1— aj  ic  — 2/     y     x 

4.  1-1.  13.    ^-  +  ?-l. 
a;     !/  a;-2/      j/     y 

6.    — ,-4t-  14.    l  +  «-^ 

a  —  b      a-j-b  x      y     z 


1 

a;-2 

1 
cc  +  l 

1 

a-h4 

2 
a-3 

1 

1 

A;-l     A;  +  l 


15.    -  +  -  + 
X     y      z 


7     _i 3  g &_ 

c-2      (c-l)(c-2)  *    (a;-l)(a;  +  3)      a;  +  3* 

8.    1-1  +  i.  iT.-i+J-H-  ^ 


a      6      c  a;+2^     x-y     (x-{-y)(x-y)' 

9.1+1.1.  18.      1    +    1_.  1 


ab     be     abc  a+1    a  —  1     (a-\-l){a—l) 


MULTIPLICATION   AND  DIVISION 
99.    Examples.     1.   Multiply  ^  by  5,  also  -  by  c. 

2.  Multiply  I  by  4,     also  —  by  c. 

6c 

3.  Multiply  I  by  f     also  ^  by  ^. 

0        a 

4.  Multiply         ^(^  +  ''^)       by        ^(^-^)       . 

Solution,  -1(^:M)_x-AJ2^^ 


(>-a)(x+2)     U(>+^(2-2)      (x  +  2)(ar-2)      x2-4 
The  first  step  is  to  cancel  all  possible  factors. 


86  SIMPLE   FRACTIONS 

EXERCISES 

Perform  the  following  multiplications : 

1.    ^x--  3.    —X  —  -  5.    ^^y  X  ^^  • 

c       a  xy      ac  nyc      mx 

ah      c^x"  A     c^^cy  ft     rst     xy^ 

cx       a  by'     dx  xy      'rs 

7.  3  ^     3c ^(c-3)(c+3), 

n  +  8      ^    ^   ^  (3_c)(3  +  c)  6cd 

8.    ? X3(a-2).10.  ^^±Jx(^^  +  ^)(^^-^\ 

(a-2)(a+3)  ^  2 +  3  a;       (a +  6) (3  a +  6) 

11     ?_o^x—         12     (a  +  ^^)(a  +  26)      (2a-6^(a-26) 
2cd      10a*  *  a-26  (3  a+ 6)(a  +  26)* 

100.    Examples.     1.    Divide  ^  by  2,   also  —  by  a. 


2.    Divide  f  by  7,  also  -  by  c. 


3.  Divide  j  by  f,  also  -  by  ^. 

4.  Divide  ; — —  by 


(a;-l)(a;-2)     ^  x-l' 

Solution.     ^ -^  -J-  = ^ X  ^=:^  =  -^ . 

(a:-l)(a;-2)      x-l      (>-^(x-2)  \         x-2 

After  inverting  the  terms  of  the  divisor  cancel  all  possible 
factors. 

EXBRCISBS 

Perform  the  following  divisions : 

-     a6  ^  ^  -     abc     be  -     mxy      ny 

c  xy      ax  my      mx 

2     —^^.  A    ^  ^9^  ft     rg^  ^  1^3 

c       c  bjf  '  cy  '   xy     Qc^y 


EQUATIONS   INVOLVING   FRACTIONS  87 

7.   J^-j-a6.  9.  «  +  l  .  3(a+l)^  ^^       4  A:^      .      2fe 

'  x  —  1  '       '  '  a  —  2  '      rt  +  4'  — 9  abc^  '       ah 

a-h     ^^^  25c         15  (a:  +  2/)'     aj  +  y 

13     6a-3  .2a-3  ^^     (2+a^)(2-a^)  .  (2-a;)^ 

5aj      '     15a^   *  '       3a;(icH-l)      *  a;(a;-2)' 


Note.  —  A  full  treatment  of  fractions  will  be  found  toward  the  end 
of  this  course.  To  give  such  a  treatment  here  would  necessitate  a 
complete  study  of  factoring  and  consequently  the  postponement  of 
much  important  work  on  the  solution  of  equations  and  problems 
which  naturally  belongs  early  in  the  course.  The  treatment  given 
in  this  chapter  is  based  directly  on  Arithmetic  and  is  suflBcient  for  the 
solution  of  all  problems  depending  on  simple  equations  which  nat- 
urally occur  in  an  elementary  course. 

EQUATIONS  INVOLVING  FRACTIONS 
Solve  n  + 1+1  =  88.  (1) 

First  Solution.     The  coefficients  of  n  are  1,  J,  and  \. 


Applying  Principle  I, 

(1+Ki)n  =  lin  =  88. 

(2) 

By  D  1  If, 

n  =  88  -^  1|  =  48. 

(3) 

Second  Solution.     Multiply  both  members  of  (1)  by  6. 

Thatisjbyilfie, 

6n  +  ?i^  +  ^  =  528. 
2         3 

(2) 

By  F,  V, 

6n  +  3n  +  2n  =  528. 

(3) 

By  F,  I, 

11  n  =  528. 

(*) 

By  D 1 11, 

n  =  48,  as  before. 

(5) 

The  object  is  to  multiply  both  members  of  the  equation  by  such 
a  number  as  will  cancel  each  denominator.  Hence  the  multiplier 
must  contain  each  denominator  as  a  factor. 

Evidently  12  or  18  might  have  been  chosen  for  this  purpose,  but 
not  8  or  10.     6  is  the  smallest  number  which  will  cancel  both  2  and  8. 


88  SIMPLE  FRACTIONS 

101.  The  process  explained  in  the  second  solution  above  is 
called  clearing  of  fractions. 

As  another  illustration  solve  the  equation 

_i 1_  = 4 

x  +  1      x-1      (x-{-l)(x-l)  ^  ^ 

Here  the  lowest  multiple  available  is  (x  +  l)(x  —  1). 

Hence     40sa)C:r-l)      (x  +  1)(>-^)  _  4C5>4a)(>^l)      .^. 

By  F,  V,  each  denominator  is  now  cancelled, 

4c(x  -  1)  -  (a:  +  1)  =  4.  (3) 

By  F,  IT,  XIT,  4  a;  -  4  -  a;  -  1  =  4.  (4) 

By  F,  I  and  ^  I  5,  Sx  =  9.  (5) 

Byi)|3,  x  =  3.  (6) 

Check.     Substitute  a:  =  3  in  (1)  and  get 

1  -  i  =  I  or  I  =  -|. 

After  a  little  practice  step  (2)  should  be  performed  mentally 
and  equation  (3)  should  be  derived  immediately  from  (1). 

EXERCISES  AND  PROBLEMS 

Solve  the  following  equations,  indicating  the  principles  used 
at  each  step.  Check  each  solution  by  substituting  the  value 
obtained  in  the  original  equation.  Translate  the  first  four 
into  problems. 

For  instance,  from  the  equation  in  Ex.  2 :  Find  a  number  such 
that  when  increased  by  its  half,  its  third,  and  its  fourth,  the  sum  is  25. 

'•2  +  3  =  ^-  3.  -  +  4»-^=_+26. 

a.  »  +  ^  +  |4-|  =  26.         4.  1  +  1-^,^  =  82. 

5.   7a;  +  i^4-^  +  23=f+~  +  5^  +  113 

7  5  D       10 

^    J,      ,6n     3w-f2,./j 

7  2 


EQUATIONS  INVOLVING  FRACTIONS  89 

3  5  3 

7  5 

_    5a  +  7  ,  2aH-4     3a  +  9  ,  p. 
^-   ~Y~+~3~"~4~+^- 

17a-5     10a  +  2^5a  +  7      ^  ?_1  =  1 

3  4  2  •  *  a;      2     a;' 

^^    5^  +  5J2^_M0)^2^^24.  13.   i  +  ^  =  i. 

2  5  3a;     2aj     6 

14.  J-  +  i  =  A  +  l.  16.1-2  8 


4a;3a;6a;2  a;a;  — 1     a;(x  — 1) 


16. 


aj  +  l      a;  — 1      (a;  +  l)(a;  — 1) 


17.^+      '  * 


1      a;  +  l      (a;-l)(a;  +  l) 

18.  The  sum  of  two  numbers  is  12,  and  the  first  number  is  ^ 
as  great  as  the  second.     What  are  the  numbers  ?  , 

19.  The  smaller  of  two  numbers  is  |  of  the  larger.  If  their 
sum  is  66,  what  are  the  numbers  ? 

20.  Find  two  consecutive  integers  such  that  ^  of  the  first 
minus  ^^  of  the  second  equals  9. 

21.  Find  three  consecutive  integers  such  that  ^  of  the  first 
plus  the  second  minus  ^  the  third  equals  5. 

22.  Find  three  consecutive  integers  such  that  ^  of  the  first 
plus  J  of  the  second  minus  J^  of  the  third  equals  28. 

23.  There  are  three  numbers  such  that  the  second  is  4  more 
than  9  times  the  first,  and  the  third  is  2  more  than  6  times  the 
first.  If  1  of  the  third  is  subtracted  from  |  of  the  second,  the 
remainder  is  3.     Find  the  numbers. 

24.  There  are  three  numbers  such  that  the  second  is  2  more 
than  9  times  the  first  and  the  third  is  5  more  than  11  times  the 
first.  The  remainder  when  ^  of  the  third  is  substracted  from 
^  of  the  second  is  one.     Find  the  numbers. 


90  SIMPLE   FRACTIONS 

25.  What  number  must  be  subtracted  from  both  the  nu- 
merator and  the  denominator  of  the  fraction  |  in  order  to 
make  the  result  equal  to  |^. 

26.  What  number  must  be  subtracted  from  both  numerator 
and  denominator  of  the  fraction  f  in  order  that  the  fraction 
may  be  increased  three-fold?     Ans.  2\. 

27.  What  number  added  to  both  numerator  and  denomina- 
tor of  the  fraction  f  will  double  the  fraction  ? 

28.  Find  a  number  of  two  digits  in  which  the  tens'  digit  is 
3  greater  than  the  units'  digit,  and  such  that  if  the  number  is 
divided  by  the  sum  of  the  digits,  the  quotient  is  7. 

29.  In  a  number  of  two  digits  the  units'  digit  exceeds  the 
tens'  digit  by  4,  and  when  the  number  is  divided  by  the  sum  of 
its  digits  the  quotient  is  4.     Find  the  number. 

30.  There  are  two  numbers  whose  sum  is  40,  such  that  if  the 
greater  is  divided  by  their  difference  the  quotient  is  3. 

31.  The  shortest  railway  route  from  Boston  to  Chicago  is  166 
miles  more  than  4  times  that  from  Boston  to  New  York  ;  and 
the  shortest  route  from  Boston  to  Atlanta  is  196  miles  less  than 
6  times  that  from  Boston  to  New  York.  The  distance  from 
Boston  to  Chicago  is  481  miles  more  than  ^  the  distance  from 
Boston  to  Atlanta.     Find  each  of  the  three  distances. 

REVIEW  QUESTIONS 

1.  What  principle  is  used  in  reducing  fractions  to  a  common 
denominator  ? 

2.  State  from  arithmetic  the  rules  for  multiplying  or  divid- 
ing a  fraction  by  an  integer ;  by  another  fraction.  State  these 
as  formulas  by  using  letters. 

3.  What  is  the  effect  on  an  equation  if  both  members  are 
multiplied  by  the  same  number?  How  may  this  be  used  to 
clear  an  equation  of  fractions  ? 

4.  Complete  your  list  of  principles  stated  in  symbolic  form 
up  to  a«d  including  Principle  XV. 


DRILL  EXERCISES  91 


DRILL  EXERCISES 

1     «,«  +  7      a-3^a  +  227      ^ 
*   3         4  3     ~       5 

2.  (a-2-3c-8  +  2&)(6-a-c-6H-8). 

3.  £:zl  +  ^^  +  ^Lz:^==2  +  a. 

2  2  12 

4.  a26-(3  6-8a2-7)  +  3a6^-(4a6^+8-2a2). 

5.  ^  +  ^  +  ^^  =  2a-26. 

4  4  4 

«    w4-l,n4-3,n-l      n  +  13,n-2 

'•  ^-^-T-  +  -T-  =  -3-  +  ^-       . 

7.  16ax-^4:-(S-Sax-d)-(12ax-lS-ax). 

8.  Add  15aa^  +  3bc^,  2  ftc^  -  7  aor^,  and  5  +  2  aar'-S  ftc^. 

9.  Add  16  -  7  a6  -  2  a^  -h  5  a6,  4  a^  -  2  a6,  and  5  a6  -  8. 

10.  Add  51  ar^2/ -  35  +  12  a^,  41  -  17  a"  -  57  ar^y,  and3a^?/. 

11.  Add  35  6^  _  13  c2,  8  c^  -  3  6V,  and  6  6^  _  8  c^-  9  6V. 

12.  Add  19  -  2  a;  +  3  ^2^  +  2  6,  4  ar^6  H-  a;  +  5  6  +  8,  and  4  a;. 

4^8     16     2^32 
14    .V-3     y4-9^y-ll      ^ 
4  10  4  • 

15.  .V^  +  ^2^.V±_8^2,-20. 

16.  |4-^  +  ^  =  25. 

17.  1-^  +  2^  +  ^  =  15. 
c)  o  o  2 

18.  ^  ^  4 


x-l      a  +  l      (a;-l)(a;  +  l) 
19.        a?-l  [  a?  +  l  27 


a;  +  2     a;-3      (a;  +  2)(a;-3) 
2Q    2a;-l      4a;-l  -10 


a;  +  2       2a;-3      (a;  +  2)(2  a:- 3) 


CHAPTER   VI 
LITERAL   EQUATIONS   AND   THEIR   USES 

102.  Some  of  the  advantages  of  algebra  over  arithmetic  in 
solving  problems  have  been  pointed  out  in  the  preceding  chap- 
ters. For  instance,  the  brevity  and  simplicity  of  statement 
secured  through  the  use  of  letters  to  represent  numbers;  the 
translation  of  problems  into  equations  ;  and  the  clear  and  logical 
solution  of  these  equations,  step  by  step. 

Another  advantage  is  set  forth  in  the  present  chapter; 
namely,  the  opportunity  offered  in  Algebra  to  summarize  the 
solution  of  a  whole  class  of  problems  by  solving  what  is  called  a 
literal  equation,  thus  obtaining  a  formula  which  may  be  used  in 
solving  other  problems. 

For  example,  in  arithmetic  we  solved  many  problems  obtain- 
ing the  interest  when  the  principal,  rate,  and  time  were  given. 
We  now  see  that  all  of  these  can  be  summarized  in  the  one 
literal  equation  /=:Dri 

Furthermore,  the  rules  for  obtaining  the  principal,  the  rate, 
or  the  time  may  now  be  derived  directly  from  this  equation 
by  Principle  VI,  thus  obtaining : 

rt'  pt^  pr 

Translate  each  of  these  formulas  into  a  rule  of  arithmetic. 

103.  Definition.     The  process  of  deriving^  =—  from  i=prt 

rt 

is  called  solving  the  equation  i—pi-t  for  p  in  terms  of  t,  r,  and 
ty  or  simply  solving  the  equation  for  p. 

92 


PROBLEMS  FROM  ARITHMETIC  93 

104.  In  arithmetic  a  problem,  is  said  to  be  solved  when  a 
numerical  answer  is  obtained  which  satisfies  the  conditions 
given.  The  solutions  thus  far  found  in  algebra  have,  for  the 
most  part,  been  of  this  sort. 

It  is  customary,  however,  to  say  that  a  problem  has  been 
solved  in  the  algebraic  sense  when  a  formula  is  fouud  which 
gives  complete  directions  for  deriving  the  numerical  answer. 

Thus,  /)  =  —  is  a  solution  for  the  principal  since  it  states  precisely 

how  to  find  the  principal  in  terms  of  interest,  rate,  and  time. 

105.  It  is  thus  seen  that  from  the  literal  equation  i  =  prt 
we  obtain  the  complete  solution  of  every  problem  which  calls 
for  any  one  of  these  four  numbers  in  terms  of  the  other 
three. 

In  modern  times  machines  are  extensively  used  for  convpu- 
tation.  The  algebraic  solution  of  a  literal  equation  gets  the 
problem  ready  for  the  computing  machine^  that  is,  it  gets  the 
formula  which  the  computer  must  use. 

LITERAL  EQUATIONS   USED  IN  SOLVING  PROBLEMS 
I.  PROBLBMS  FROM  ARITHMETIC 

1.  If  $700  is  invested  at  5%  simple  interest,  what  is  the 
amount  at  the  end  of  5  years  ?  This  problem  calls  for  the 
amount,  which  is  the  sum  of  principal  and  interest. 

If  a  =  amount,  then  a—p-\-i  —  p-\- pri  =  yo(l  +ri). 

Applying  this  formula,  a  —  700(1  -f  -^  •  5)  =  875. 

2.  Solve  the  equation  a=^p  -^-prt  for  p  in  terms  a,  r,  and  t. 
State  a  rule  of  arithmetic  represented  by  this  solution  and  make 
a  problem  which  can  be  solved  by  means  of  it. 

3.  Solve  a  =])-{- prt  for  r  in  terms  of  a,  p,  and  t.  State  a 
rule  of  arithmetic  represented  by  this  solution  and  make  a  problem 
which  can  be  solved  by  means  of  it. 


94  LITERAL   EQUATIONS  AND   THEIR    USES 

4.  Solve  a  =  p-\-prt  for  t  in  terms  of  a,  p,  and  r.  State  a 
rule  of  arithmetic  represented  by  this  solution  and  make  a 
problem  which  can  be  solved  by  means  of  it. 

5.  A  real  estate  dealer  sold  a  house  and  lot  for  $  7500,  for 
which  he  received  a  commission  of  4  %.     What  was  his  profit? 

Solution.     Letting  c  =  commission,  b  =  base,  and  r  =  rate,  we  have 

c  =  br. 
Applying  this  formula,  c  =  br  =  7500  •  j-^^  =  75  •  4  =  300. 

6.  Solve  the  equation  c=br  for  b  in  terms  of  c  and  r. 

7.  Solve  c  =  hr  for  r  in  terms  of  c  and  b. 

8.  State  the  rules  of  arithmetic  represented  by  the  solutions 
in  Exs.  6  and  7,  and  make  problems  to  be  solved  by  these  rules. 

9.  How  much  must  I  remit  to  my  broker  in  order  that  he 
may  buy  $  600  worth  of  bonds  and  reserve  5  %  commission  ? 

]*must  send  him  both  base  and  commission.  Calling  this  the  amount 
and  representing  it  by  a,  we  have 

a  =  b  +  c  =  b  +  br  =  b(l-\-r). 

Hence,  a  =  600  (1  +  ib)  =  630. 

10.  Solve  a  =  b-{-br  for  b  and  translate  the  result  into  words. 

11.  Solve  a  =  b  -\-br  for  r  and  translate  the  result  into  words. 

12.  A  dealer  sold  berries  for  $  18.95,  and  after  deducting  a 
commission  of  2%  sent  the  balance  to  the  truck  gardener. 
How  much  did  he  remit  ? 

The  sum  he  sent  was  the  difference  between  the  base  and  the  com- 
mission ;  calling  this  d,  we  have 

d=b-c==b-br=b{l-r). 

Hence,  in  this  case  d  =  18.95  (1  -  tItf)  =  18-57. 

13.  Solve  the  equation  d  =  b  (1  —  r)  for  b  in  terms  of  d  and  r. 

14.  Solve  the  equation  d  =  b  —  br  for  r  in  terms  of  b  and  d. 

15.  State  the  rules  of  arithmetic  represented  by  the  solutions 
in  Exs.  13  and  14  and  make  problems  to  be  solved  by  these  rules. 


PROBLEMS   INVOLVING  MOTION  95 

II.     PROBLEMS  INVOLVING  MOTION 

106.  In  scientific  language  the  distance  passed  over  by  a 
moving  body  is  called  the  space,  and  the  number  of  units  of 
space  traversed  is  represented  by  s.  The  rate  of  uniform  mo- 
tion, that  is,  the  number  of  units  of  space  traversed  in  each 
unit  of  time,  is  called  the  velocity,  and  is  represented  by  v. 
The  number  of  units  of  time  occupied  is  represented  by  t. 

Ex.  1.  If  a  train  runs  40  miles  per  hour,  how  far  does  it  run 
in  5  hours  ? 

Ex.  2.  At  a  certain  temperature  sound  travels  1080  feet  per 
second.     How  far  does  it  travel  in  5  seconds  ? 

In  each  of  these  examples  the  space  passed  over  is  found  by  multi- 
plying the  velocity  by  the  time.     Using  the  symbols  s,  v,  and  t,  we 

^^""^  S  =  Iff.  (1) 

1.  Solve  the  equation  s  =  vt  for  t  in  terms  of  s  and  v,  and 
for  V  in  terms  of  s  and  t. 

Translate  each  of  these  formulas  into  words. 

It  is  to  be  understood  in  all  problems  here  considered  that  the  velocity 
remains  the  same  throughout  the  period  of  motion ;  e.g.  sound  travels 
just  as  far  in  any  one  second  as  in  any  other  second  of  its  passage. 

2.  If  sound  travels  1080  feet  per  second,  how  far  does  it 
travel  in  6  seconds  ? 

3.  If  a  transcontinental  train  averages  35  miles  per  hour, 
how  far  does  it  travel  in  2J  days  ?  (Given  v  =  35,  t  =  2^  -  24, 
to  find  s.) 

4.  A  hound  runs  23  yards  per  second  and  a  hare  21  yards 
per  second.  If  the  hound  starts  79  yards  behind  the  hare, 
how  long  will  it  require  to  overtake  the  hare  ? 

If  t  is  the  number  of  seconds  required,  then  by  formula  (1)  during 
this  time  the  hound  runs  23  t  yards  and  the  hare  runs  21 1  yards.  Since 
the  hound  must  run  79  yards  farther  than  the  hare,  we  have: 
23t  =  21t  +  79. 


96  LITERAL  EQUATIONS   AND   THEIR   USES 

5.  An  ocean  liner  making  21  knots  an  hour  leaves  port  when 
a  freight  boat  making  8  knots  an  hour  is  already  1240  knots 
out.     In  how  long  a  time  will  the  liner  overtake  the  freight  ? 

6.  A  motor  boat  starts  7f  miles  behind  a  sailboat  and  runs 
11  miles  per  hour  while  the  sailboat  makes  61-  miles  per  hour. 
How  long  will  it  require  the  motor  boat  to  overtake  the  sailboat  ? 

7.  A  freight  train  running  25  miles  an  hour  is  200  miles 
ahead  of  an  express  train  running  45  miles  an  hour.  How  long 
before  the  express  will  overtake  the  freight  ? 

8.  A  bicyclist  averaging  12  miles  an  hour  is  52  miles  ahead 
of  an  automobile  running  20  miles  an  hour.  How  soon  will  the 
automobile  overtake  him  ? 

9.  A  and  B  run  a  mile  race.  A  runs  18  feet  per  second 
and  B  17|^  feet  per  second.  B  has  a  start  of  30  yards.  In 
how  many  seconds  will  A  overtake  B  ?  Which  will  win  the 
race  ? 

If  in  each  of  the  examples  4  to  9  we  oall  the  velocity  of  the 
first  moving  object  Vi  (read  v  one)  and  that  of  the  second  V2 
(read  v  two),  then  the  distance  traveled  by  the  first  in  the  re- 
quired time  t  is  Vity  and  that  traveled  by  the  second  is  V2t. 

Then  if  n  is  the  distance  which  the  second  must  go  in  order 
to  overtake  the  first,  we  have 

V2t=v,t  +  n.  (2) 

The  solution  of  (2)  for  t  gives  the  time  required  in  each  problem 
for  the  second  to  overtake  the  first. 

Equation  (2)  summarizes  the  solution  of  all  problems  like  those 
from  4  to  9. 

It  is  important  that  formulas  (1)  and  (2)  be  clearly  under- 
stood, since  they  are  used  very  often  in  problems  on  motion. 

10.  A  fleet,  making  11  knots  per  hour,  is  1240  knots  from 
port  when  a  cruiser,  making  19  knots  per  hour,  starts  out  to 
overtake  it.     How  long  will  it  require  ? 

Use  formula  (2). 


PROBLEMS  INVOLVING  MOTION  97 

11.  In  how  many  minutes  does  the  minute  hand  of  a  clock 
gain  15  minute  spaces  on  the  hour  hand  ? 

Using  one  minute  space  for  the  unit  of  distance  and  1  minute  as 
the  unit  of  time,  the  rates  are  1  and  ^^  respectively,  since  the  hour  hand 
goes  jij  of  a  minute  space  in  1  minute.  Letting  i  be  the  number  of 
minutes  required,  we  have,  using  formula  (2), 
l-t  =  ^^t  +  15. 

12.  In  how  many  minutes  after  4  o'clock 
will  the  hour  and  minute  hands  be  together  ? 
(Here  the  minute  hand  must  gain  20  minute 
spaces.)     Ans.  21-j\  min. 

13.  At  what  time  between  5  and  6  o'clock  is  the  minute 
liand  15  minute  spaces  behind  the  hour  hand  ?  At  what  time 
is  it  15  minute  spaces  ahead  ? 

Since,  at  5  o'clock,  it  is  25  minute  spaces  behind  the  hour  hand,  in 
the  first  case  it  must  gain  25  —  15  =  10  minute  spaces,  and  in  the 
second  case  it  must  gain  25  -f  15  =  40  minute  spaces.  Make  a  dia- 
gram as  in  the  preceding  problem  to  show  both  cases. 

14.  At  what  time  between  9  and  10  o'clock  is  the  minute 
hand  of  a  clock  30  minute  spaces  behind  the  hour  hand  ?  At 
what  time  are  they  together  ? 

In  each  case,  starting  at  9  o'clock,  how  much  has  the  minute  hand 
to  gain  ? 

15.  A  fast  freight  leaves  Chicago  for  New  York  at  8.30  a.m., 
averaging  32  miles  per  hour.  At  2.30  p.m.  a  limited  express 
leaves  Chicago  over  the  same  road,  averaging  55  miles  per  hour. 
In  how  many  hours  will  the  express  overtake  the  freight  ? 

If  the  express  requires  t  hours  to  overtake  the  freight,  the  latter 
had  been  on  the  way  t  +  Q  hours.  Then  the  distance  covered  by  the 
express  is  55  <,  and  the  distance  covered  by  the  freight  is  32  (f  +  6). 
As  these  must  be  equal,  we  have  bot  =  'd2(t  +  6). 

16.  In  a  century  bicycle  race  one  rider  averages  19|-  miles 
per  hour,  while  another,  starting  40  minutes  later,  averages 
22^  miles  per  hour.  In  how  long  a  time  will  the  latter  over- 
take the  former  ? 


98  LITERAL   EQUATIONS  AND  THEIR   USES 

III.     PROBLEMS  INVOLVING  THE  LEVER 

107.   Two  boys,  A  and  B,  play  at  teeter.     They  find  that  the 

teeter  board  will  balance  when  equal  products  are  obtained  by 

multiplying  the  weight  of  each  by  his  distance  from  the  point 

of  support.  ^,        .,  -_       •  u     on 

^^  Thus,  if  B  weighs  80 

^(100 lbs.) BisoihB^  pounds    and    is    5   feet 

-^^  -4.  feet,  "  fifeer  from  the  point  of  sup- 

port, then  A,  who  weighs  100  pounds,  must  be  4  feet  from  this  point, 
since  80  x  5  =  100  x  4. 

The  teeter  board  is  a  certain  kind  of  lever;  the  point  of 
support  is  called  the  fulcrum. 

In  each  of  the  following  problems  make  a  diagram  similar 
to  the  above  figure : 

1.  A  and  B  weigh  90  and  105  pounds  respectively.  If  A  is 
seated  7  feet  from  the  fulcrum,  how  far  is  B  from  that  point  ? 

2.  Using  the  same  weights  as  in  the  preceding  problem,  if 
B  is  6|  feet  from  the  fulcrum,  how  far  is  A  from  that  point  ? 

3.  A  and  B  are  5  and  7  feet  respectively  from  the  fulcrum. 
If  B  weighs  75  pounds,  how  much  does  A  weigh  ? 

4.  A  and  B  weigh  100  and  110  pounds  respectively.  A 
places  a  stone  on  the  board  with  him  so  that  they  balance 
when  jB  is  6  feet  from  the  fulcrum  and  A  5i  feet  from  this 
point.     How  heavy  is  the  stone  ? 

5.  If  the  distances  from  the  boys  to  the  fulcrum  are  respec- 
tively di  and  dz,  and  their  weights  Wi  and  W2,  then 

</iiVi  =  (/2IV2.  (1) 

This  equation  is  a  statement  in  the  language  of  algebra  of  a  very 
important  law  of  nature.  The  law  is  the  result  of  a  very  large  num- 
ber of  careful  experiments.  It  is  a  universal  custom  among  scientific 
men  so  far  as  possible  to  express  laws  of  nature  by  means  of  literal 
equations  of  this  sort. 

If  any  three  of  the  four  numbers  r/j,  m;,,  do,  w^,  are  given,  the  fourth 
may  be  found  by  means  of  the  equation  d^w^  =  </2"^2' 


PROBLEMS  INVOLVING  THE  LEVER         99 

6.  Solve  diiVi  =  d^Wc^  for  each  of  the  four  numbers  involved 
in  terms  of  the  other  three. 

7.  A  and  B  are  seated  at  the  opposite  ends  of  a  13-foot 
teeter  board.  Using  the  weights  of  problem  1,  where  must  the 
fulcrum  be  located  so  that  they  shall  balance? 

If  the  fulcrum  is  the  distance  d  from  A,  then  it  is  (13  —  d)  from  B. 
Hence,  90  rf  =  105(13  -  d). 

8.  A^  who  weighs  75  pounds,  sits  7  feet  from  the  fulcrum, 
and  5,  who  weighs  105  pounds,  sits  on  the  other  side.  At 
what  distance  from  the  fulcrum  should  B  sit  in  order  to  make 
a  balance  ? 

9.  A  and  B  together  weigh  2121  pounds.  They  balance 
when  ^  is  6  feet,  and  B  6|  feet,  from  the  fulcrum.  Find  the 
weight  of  each. 

10.  A  lever  9  feet  long  carries  weights  of  17  and  32  pounds 
at  its  ends.  Where  should  the  fulcrum  be  placed  so  as  to  make 
the  lever  balance  ? 

11.  A  lever  of  unknown  length  is  balanced  when  weights  of 
30  and  45  pounds  are  placed  on  it  at  opposite  ends.  Find  the 
length  of  the  lever,  if  the  smaller  weight  is  two  feet  farther 
from  the  fulcrum  than  the  greater. 

Suggestion.  Let  x  be  the  distance  from  the  greater  weight  to  the 
fulcrum. 

EXERCISES 

Solve  the  following  for  each  letter  in  terms  of  the  others : 

1.  1^=32+ I  C.  3. 

2.  l  —  a-\-{n  —  V)d.  4. 

Solve  each  of  the  following  for  x : 
h.   ax-\-Z'b  =  cx-\-d.  g 

6.  {a—x){b-\-x^=.x{b  —  x). 

7.  {x-\-a){x-\-h)^{x-cf.         ^' 


=  |(«  +  0- 

a-rl 

ax+b      bx  +  c_^ 
c              d 

X         X     _     a 
b     a—  b     a  +  6 

100  LITERAL   EQUATIONS   AND  THEIR   USES 

x-{-l      a  +  b 


10. 


1     a-b 


11.  f +  ±  +  :f  =  l. 
a     b     c 


12. 


13. 


a-j-bx  _  c-[-dx 
a-\-b        c-^d 

a  +  b  _a— b 
1  H-rc     1  —X 


14. 

Sax 

a-b 

-2a  =  5 

X. 

15. 

ax—  1 
bx 

_l-^bx 
ax 

+  1. 

16. 

x  —  a 
a-b 

x-i-a 
a  +  b 

2x 

a-\-b 

17. 

a  +  x 

b-\-x     c 

■i-x  ■ 

a 

b 

c 

2x( 

x  +  1) 

18     ^il-L^Zll 

a;-l      ic  +  l      (x-{-l)(x-l) 

-^    x-\-m      x  —  m  _      2  x{x  -\- 1) 
x  —  m     x-\-m     {x -\- m){x  —  m) 


20. 


a;  +  &  1^  — <^_     2a;(a;  +  a) 
x  —  a     x-\-a      {x  —  a){x-\-a) 


21.    _?^+      « 


REVIEW  QUESTIONS 

1.  Make  a  list  of  all  the  rules  for  interest  which  are  derived 
from  i=prt. 

2.  Make  a  list  of  all  the  rules  for  percentage  which  are  de- 
rived from  a=p-\- prt. 

3.  Make  a  list  of  the  rules  for  commission  which  may  be 
obtained  from  c  —  br,a  =  b-\-c  =  b-{-br,  d  =  b  —  c  =  b  —  br. 

4.  What  problems  on  motion  in  this  chapter  belong  to  the 
class  whose  solutions  are  summarized  by  the  solutions  of  the 
equation,  ^^  =  y^t  ^  ^^ 

5.  State  fully  the  meaning  of  the  equation  Widi  =  ^3(^2  ^^ 
connection  with  the  lever. 

6.  What  is  the  difference  in  meaning  between  the  sokltion 
of  a  special  problem  in  arithmetic  and  that  of  a  problem  in- 
volving a  literal  equation  in  algebra? 


DRILL   EXERCISES  101 

DRILL  EXERCISES 

1.  Subtract  2a-6  a^b-3x-\-21  from  19  -2x-\-Sx'b-7a. 

2.  From  6a- 45 +  86-3c  + 82  c&  subtract  7  6  + 18  +  6  c. 

3.  (17H-2a-36-4c)(2-a  +  6-c). 

4.  (13c-4d  +  8e-3)(c-rf). 

5.  (4.xy-2y-3x-2){y-x-\-y  +  5). 

6.  (9ax-Sx~5a-2x-\-4:){5-x). 

7.  (9a;-3)(4-a^)  +  (a:-3)2  =  -8(a:  +  2)2-f 94. 

8.  {x  +  1)2+ (a;  +  2)2+ (a;  +  3)=*=  (3  x  -  l)(x  + 12)  -  43. 

9.  (2a;+5)(a;-7)-(a;-l)2  =  (a;  +  l)(a;  +  2)-28. 

10.  3(5  -xy~(2x-  l)(x -  1)  =  (x -  7){x  + 10)+ 17  x  +  50. 

11.  (32  +  a;)(4  a;  - 1)  +  (5  -  xf-^-  {x  -  If  =  6(a;  + 1)^  + 194. 

12.  (2aj-7)(5-a;)-(2-5.T)(l-a;)  =  -a:(7a;-34)-17. 
a;  +  8     a;-9      a;-17^4a;-7      2a;  +  6      5-31a; 

'       2  12    "^6  2"^312' 

14    3a;-l      3a;  +  3      a;-l^a;  +  5      ^         20 
6  3  2  6  3 

15.   ^Z1^  +  ^±J?=2. 
a  6 

,^    3a;-16  ,     21        6a;-ll 

lo. i = • 

2  a:-8  4 

17.  If  two  numbers  differ  by  d  and  if  the  greater  of  the 
numbers  is  x,  how  do  you  represent  the  other  ? 

18.  A  father  is  3  times  as  old  now  as  his  son  was  7  years 
ago.  If  the  son's  age  now  is  represented  by  x,  how  is  the 
father's  age  represented  ? 

19.  A  picture  inside  the  frame  is  w  inches  wide  and  w  +  6 
inches  long.  The  frame  is  4  inches  wide.  Express  the  area 
of  the  frame  in  terms  of  ic. 

20.  A  picture  inside  the  frame  is  w  inches  wide  and  I  inches 
long.  The  frame  is  a  inches  wide.  Express  the  area  of  the 
frame  in  terms  of  a,  w,  and  I. 


CHAPTER   VII 
GRAPHIC   REPRESENTATION 

108.  Graphic  Representation  of  Statistics.  A  graphic  repre- 
sentation of  the  temperatures  recorded  on  a  certain  day  is 
shown  on  the  next  page.     The  readings  were  as  follows : 


3  P.M. 

29° 

9  P.M. 

21° 

3  a.m. 

12° 

9a.m.     12° 

4  P.M. 

29° 

10  P.M. 

20° 

4  a.m. 

11° 

10  A.M.      13° 

5  P.M. 

28° 

•    11p.m. 

17° 

5  a.m. 

10° 

11  A.M.      16° 

6  P.M. 

26° 

12  m't. 

16° 

6  a.m. 

10° 

12  Noon  17° 

7  P.M. 

24° 

1  a.m. 

14° 

7  A.M. 

10° 

1  P.M.      18° 

8  p.m. 

22° 

2  a.m. 

12° 

8  A.M. 

10° 

2  P.M.     20° 

In  the  graph  each  heavy  dot  represents  the  temperature  at  a  cer- 
tain hour.  The-  distance  of  a  dot  to  the  right  of  the  heavy  vertical 
line  indicates  the  hour  of  the  day  counted  from  noon,  and  its  distance 
above  the  heavy  horizontal  line  indicates  the  thermometer  reading  at 
that  hour.  The  lines  joining  these  dots  complete  the  picture  repre- 
senting the  gradual  changes  of  temperature  from  hour  to  hour. 

Graphs  of  this  kind  are  used  in  commercial  houses  to  represent 
variations  of  sales,  fluctuations  of  prices,  etc.  They  are  used  by  the 
historian  to  represent  changes  in  population,  fluctuations  in  mineral 
productions,  etc.  In  algebra  they  are  used  in  solving  problems  and 
in  helping  to  understand  many  difficult  processes.  In  the  succeeding 
exercises  the  cross-ruled  paper  is  essential. 

Make  a  graphic  representation  of  the  tables  of  data  on  the 
opposite  page : 

In  each  case  the  number  to  be  represented  by  one  space  on  the  cross- 
ruled  paper  should  be  chosen  so  as  to  make  the  graph  go  conveniently  on  a 
sheet.  Thus  in  Ex.  1  let  one  small  horizontal  space  represent  two  years  and 
one  vertical  space  a  million  of  population ;  and  in  Ex.  3  let  one  horizontal 
space  represent  one  year  and  one  large  vertical  space  one  hundred  thousand 
of  population. 

102 


GRAPHIC   REPRESENTATION 


103 


±'vL- 

~^   St 

^ 

_^ 

■h"'}'              K 

^ 

\ 

>v 

-•■"0'                                                   >1                                                                                                                 M 

\_                                                     / 

s                           \                                     t 

3                                 ^e                                    i 

<u                                                            \                                                          J^ 

J2 .d  .1  r"^      ^                                                 V                                                          i  1_ 

-                                                                         5^                                                 / 

•    1                              V                 ^t 

i                                       -t,-             ^^ 

"^                     i     7     " 

i-10'                                                    -SXh2- 

-tjt2 

_2 

12  M.       JP-M*         6              ^             12        SA.M                         i             12     P.JM. 

TimeLihe 

EXERCISES 

1.  The   population  of  the   United  States  as  given  by  the 
census  reports  from  1800  to  1910 : 

1800  .  .    4.3  (million)  1840  .  .  17.1 

1810  .  .    7.2  1850  .  .  23.2 

1820  .  .     9.6  1860  .  .  31.4 

1830  .  .  12.9  1870  .  .  38.6 

2.  The  population  of  the  boroughs  now  constituting  greater 
New  York  City : 

1800  .  .     79  (thousand)  1840  .  .     391  1880  .  .  1912 

1810  .  .  119  1850  .  .     696  1890  .  .  2507 

1820  .  .  152  1860  .  .  1175  1900  .  .  3427 

1830  .  .  242  1870  .  .  1478  1910  .  .  4767 


1880  . 

.  50.2 

1890  . 

.  62.6 

1900  . 

.  76.3 

1910  . 

.  92.0 

104  GRAPHIC   REPRESENTATION 

3.  The  population  of  Chicago  since  1850 : 

1850  .  .     30  (thousand)    1880  .  .     503  1900  .  .  1698 

1860  .  .  109  1890  .  .  1100  1910  .  .  2185 

1870  .  .  306 

4.  Observe  the  weather  reports  in  a  daily  paper  and  make 
a  graph  representing  the  hourly  change  of  temperature  for 
twenty-four  hours. 

5.  From  your  own  state,  city,  or  town  obtain  data  which 
you  can  represent  by  means  of  graphs. 

GRAPHIC   REPRESENTATION    OF  MOTION 

109.  A  useful  picture  of  the  distance  traversed  by  a  moving 
body  can  be  made  by  a  graph  similar  to  the  preceding. 

E.g.  Suppose  a  man  is  walking  3  miles  per  hour.  We  mark  units  of 
time  from  the  starting  point  to  the  right  along  the  horizontal  reference 
line,  and  indicate  miles  traveled  by  the  number  of  units  measured  ver- 
tically upward  from  this  line.     (See  the  figure  on  the  opposite  page.) 

In  the  figure  each  horizontal  space  represents  1  hour,  and  each  ver- 
tical space  3  miles.  Then  in  1  hour  he  goes  3  miles;  in  5  hours, 
15  miles;  in  10  hours,  30  miles;  etc.  The  dots  representing  the 
distances  are  found  to  lie  on  a  straight  line. 

The  graph  shows  at  a  glance  the  answers  to  such  questions  as: 
How  many  miles  does  he  travel  in  4  hours?  in  13  hours?  How  long 
does  it  take  him  to  go  18  miles?  23  miles? 

Again,  suppose  24  hours  later  a  second  man  starts  out  on  a  bicycle 
to  overtake  the  first  man,  and  travels  9  miles  an  hour.  The  line 
drawn  from  the  24-hour  point  shows  the  distance  the  wheelman  trav- 
els in  any  number  of  hours,  counting  from  his  time  of  starting.  The 
points  marked  in  this  line  show  how  far  he  has  gone  in  1,  2,  3,  4,  5,  6 
hours,  etc.,  namely,  9,  18,  27,  36,  45,  54,  etc. 

The  point  where  these  two  lines  intersect  shows  in  how  many  hours 
after  starting  the  pedestrian  is  overtaken,  and  also  how  far  he  has  gone. 

In  like  manner  solve  the  following  by  means  of  graphs,  and 
in  each  case  suggest  other  questions  which  may  be  answered 
from  the  graph : 


GRAPHIC   REPRESENTATION  OF   MOTION 


105 


1.  A  starts  for  a  town  12  miles  distant,  walking  3  miles  per 
hour.  1^  hours  later  B  starts  for  the  same  place,  driving  7^ 
miles  per  hour.  When  does  B  overtake  A  ?  Where  is  A  when 
B  reaches  town  ? 

2.  In  a  mile  race  A  runs  6  yards  per  second,  and  B  5  yards 
per  second.  B  has  a  start  of  250  yards.  Who  will  win  the 
race  ?     How  far  in  the  lead  is  the  winner  at  the  finish  ? 

3.  In  a  century  bicycle  race  A  averages  17  miles  per  hour ; 
B,  who  starts  20  minutes  later,  averages  19  miles  per  hour.  In 
how  many  hours  will  B  overtake  A  ?  Who  will  win  the  race 
and  where  will  the  loser  be  when  the  winner  finishes? 

Let  one  vertical  small  space  represent  2  miles,  and  one  horizontal 
space  represent  6  minutes. 


106 


GRAPHIC   liEPllESENTATION 


110.  Illustrative  Problem.  A  man  rides  a  bicycle  into  the 
country  at  the  rate  of  8  miles  per  hour.  After  riding  a  certain 
distance  the  wheel  breaks  down,  and  he  walks  back  at  the  rate 
of  3  miles  per  hour.  How  far  does  he  go  before  the  accident, 
if  he  reaches  home  11  hours  after  startins:  ? 


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In  this  graph  each  large  horizontal  space  represents  1  hour,  and 
each  small  vertical  space  represents  1  mile.  The  problem  is  solved 
as  follows : 

(1)  Construct  the  line  representing  the  outward  journey  at  the 
rate  of  8  miles  per  hour,  extending  this  line  indefinitely. 

(2)  Beginning  at  the  point  corresponding  to  11  hours,  find  the 
points  representing  his  position  at  each  preceding  hour.  The  line 
connecting  these  points  represents  the  homeward  journey  at  the  rate 
of  3  miles  an  hour.  Extend  thiS  line  until  it  meets  the  first  line. 
The  point  where  the  lines  meet  represents  3  hours  and  24  miles, 
which  is  the  answer  required  in  the  problem. 


PROBLEMS 

Solve  the  following  problems  by  means  of  graphs.  In  each 
case  prepare  a  list  of  questions  which  may  be  answered  from 
the  graph. 


GRAPHIC   REPRESENTATION   OF   MOTION  107 

1.  A  man  rows  18  miles  per  hour  down  a  river,  and  2  miles 
per  hour  returning.  How  far  down  the  river  can  he  go  if  he 
wishes  to  return  in  10  hours  ? 

2.  A  man  goes  from  Chicago  to  Milwaukee  on  a  train  run- 
ning 42^  miles  per  hour,  and  returns  immediately  on  a  steamer 
going  17  miles  per  hour.  Find  the  distance,  if  the  round  trip 
requires  7  hours. 

3.  A  pleasure  trip  from  New  York  to  Atlanta  by  steamer, 
and  return  by  rail,  occupied  77  hours.  Find  the  distance,  if 
the  rate  going  was  16  miles  per  hour  and  returning  40  miles 
per  hour. 

Let  one  small  horizontal  space  represent  one  hour  and  one  small 
vertical  space  16  miles. 

4.  A  invests  $1000  at  5%,  and  B  invests  $5000  at  4  %. 
In  how  many  years  will  the  amount  (principal  and  interest) 
of  ^'s  investment  equal  the  interest  on  B^s  investment  ? 

Let  one  large  horizontal  space  represent  one  year  and  one  small 
vertical  space  $50.  Then  the  line  representing  A's  amount  starts  at 
the  point  marked  f$1000,  and  rises  one  small  vertical  space  each  year. 
The  line  representing  JB's  interest  starts  at  the  zero  point  and  rises 
four  small  vertical  spaces  each  year. 

5.  In  how  many  years  will  the  interest  on  $  6000  equal  the 
amount  on  $  2000  if  both  are  invested  at  5  %  ? 

6.  A  invests  $500  at  6  %  and  B  invests  $1000  at  5  %.  In 
how  many  years  will  ^'s  interest  differ  by  $  300  from  B^s  ? 

Let  one  small  vertical  space  represent  $  20.  In  this  case  both  lines 
start  from  the  zero  point.  Find  the  point  on  one  line  which  is  three 
large  spaces  vertically  above  the  corresponding  point  on  the  other  line. 

7.  A  freight  train  starts  from  New  York  to  Boston  averag- 
ing 30  miles  per  hour.  Three  hours-  later  a  passenger  train, 
averaging  50  miles  per  hour,  starts  from  New  York  in  the 
same  direction.  How  long  will  it  require  the  latter  to  overtake 
the  freight? 


108 


GRAPHIC   REPRESENTATION 


GRAPHIC  REPRESENTATION  OF   EQUATIONS 

111.  In  all  the  graphs  thus  far  constructed  two  lines  at  right 
angles  to  each  other  have  been  used  as  reference  lines.  These 
lines  are  called  axes.  The  location  of  a  point  in  the  plane  of 
such  a  pair  of  axes  is  completely  described  by  giving  its  dis- 
tance and  direction  from  each  of  the  axes.  The  direction  to 
the  right  of  the  vertical  axis  is  denoted  by  a  positive  sign,  and 
to  the  left,  by  a  negative  sign ;  while  direction  upward  from 
the  horizontal  axis  is  positive,  and  downward,  negative. 


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GRAPHIC  REPRESENTATION  OF  EQUATIONS    109 

The  horizontal  line  is  usually  called  the  jr-axis  and  the  ver- 
tical line  the  /-axis.  The  perpendicular  distance  of  any  point 
P  from  the  2/-axis  is  called  the  abscissa  of  the  point,  and  its 
distance  from  the  a^axis  is  called  its  ordinate.  The  abscissa 
and  ordinate  of  a  point  are  together  called  its  coordinates. 

E.g.  the  abscissa  of  point  P  in  the  above  figure  is  3  and  its  ordi- 
nate 2,  or  we  may  say  the  coordinates  of  P  are  3  and  2,  and  indicate 
it  thus :  P :  (3,  2),  writing  the  abscissa  first.  In  like  manner  for 
the  other  points  we  write  Q  :  (-  1,  3),  i?  :  (-  2,  0),  ^ :  (-  3,  -  4), 
and  r:(2,  -3). 

We  see  that  in  this  manner  every  point  in  the  plane  corresponds 
to  a  pair  of  numbers,  and  that  every  pair  of  numbers  corresponds  to 
a  point.  This  scheme  of  locating  points  by  two  reference  lines  i& 
already  familiar  to  the  pupil  in  geography,  where  cities  are  located 
by  latitude  and  longitude ;  that  is,  by  degrees  north  or  south  of  the 
equator  and  east  or  west  of  the  meridian  of  Greenwich. 

EXERCISES 

1.  With  any  convenient  scale,  locate  the  following  points: 
(2,  6),  (-3,  5),  (0,  1),  (1,  0),  (0,  0),  (0,  -1),  (0,  -5), 
(-5,0),  (21,54),  (-4,-8),  (3,  -10),  (-10,3). 

2.  Locate  the  following  series  of  points  and  then  see  if  a 
straight  line  can  be  drawn  through  them ;  (0,  0),  (1,  1),  (2,  2), 
(3,3),  (4,4),  (-1,  -1),  (-2,  -2),  (-3,  -3).  Name  still 
other  points  lying  on  the  same  line. 

3.  Locate  the  following  and  connect  theiij  by  a  line :  (t,  0), 
(1,  2),  (1,  3),  (1,  4),  (1,  5),  (1,  -2),  (1,  -3),  (1,  -4),  (1,  -5). 
Name  other  points  in  this  line. 

4.  Draw  the  line  every  one  of  whose  points  has  its  hori- 
zontal distance  —  2,  also  the  line  every  one  of  whose  points  has 
its  vertical  distance  +  3. 

5.  Locate  the  following  points  and  see  if  a  straight  line  can 
be  passed  through  them:  (1,  0),  (0,  1),  (2,  -1),  (3,  -2), 
(4,  -  3),  (- 1,  4-  2),  (-  2,  3),  (-  3,  4),  (-  4,  5),  (i,  \),  (i,  f), 
(I,  4).     Can  you  name  other  points  on  this  line  ? 


no 


GRAPHIC    REPRESENTATION 


112.  In  the  preceding  exercises,  in  certain  cases,  a  series  of 
points  has  been  found  to  lie  on  a  straight  line,  as  in  examples 
6,  7,  and  8.  Evidently  this  could  not  happen  unless  the  points 
were  located  according  to  some  definite  scheme  or  law. 

Illustrative  Problem.  Locate  a  series  of  points  whose  coor- 
dinates are  values  of  x  and  y  which  satisfy :  3  a;  +  4  2/  =  12. 

We  see  that  a:  =  0,  y  =  3,  also  a:  =  4,  y  =  0  are  pairs  of  such  values. 
Evidently  as  many  pairs  of  values  as  we  please  may  be  found  by 
giving  any  value  to  x  and  then  solving  the  equation  to  find  the  corre- 
sponding value  of  y.    A  table  may  thus  be  constructed  as  follows : 
^  Let  x  =  0,  4,      8,     12,  -4,-8,  etc. 

.    Then  find  y  =  3,  0,  -  3,  -  6,      6,       9,  etc. 


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These  pairs  of  values  for  x  and  y  correspond  to  the  points  as 
plotted  in  the  figure,  and  they  are  found  to  lie  on  a  straight 
line.     This  line  is  called  the  graph  of  the  equation. 


GRAPHIC   REPRESENTATION   OF   EQUATIONS  111 

Let  the  student  find  other  pairs  of  numbers  which  satisfy  this 
equation,  and  see  if  the  corresponding  points  lie  on  this  line.  Also 
find  the  numbers  which  correspond  to  any  chosen  point  on  this  line 
and  see  whether  they  satisfy  the  equation. 

113.  We  may  think  of  the  point  whose  coordinates  are  x  and 
y  as  moving  along  the  line  in  the  graph.  Then  both  x  and  y 
will  be  constantly  changing  or  varying,  but  subject  to  the 
relation  3  a;  +  4  2/  =  12.  x  and  y  are  therefore  called  variables, 
and  the  fixed  relation  according  to  which  they  vary  is  called  a 
functional  relation. 

One  of  the  graphs  on  page  105  represents  the  progress  of  a 
man  walking  3  miles  per  hour.  As  the  time  varies  the  point 
moves  along  the  line,  and  the  distance  is  seen  to  vary  as  the 
time  varies ;  that  is,  s  varies  with  t  according  to  the  relation 
s  =  3  f.  The  variable  s  is  said  to  be  a  function  of  the  variable  t 
because  it  is  connected  with  ^  by  a  definite  relation. 

In  the  equation  s  =  3  i  the  variable  s  increases  as  the  variable 
t  increases.  This  may  also  be  seen  directly  from  the  graph. 
In  3  a;  -f-  4  y  =  12,  y  '\&  seen  to  decrease^  as  x  increases. 

114.  Definitions.  An  equation  is  said  to  be  of  the  first  degree 
in  x  and  y  if  it  contains  each  of  these  letters  in  such  a  way  that 
neither  x  nor  y  is  multiplied  by  itself  or  by  the  other. 

E.g.  13  a:  —  5 y  =  14  is  of  the  first  degree,  while  2xy  —  x  =  ^  and  3  x 
—  5  y2  =  13  are  not  of  the  first  degree  in  x  and  y. 

Every  equation  of  the  first  degree  in  two  variables  has  for 
its  graph  a  straight  line ;  hence  such  an  equation  is  commonly 
called  a  linear  equation. 

115.  To  graph  an  equation  of  the  first  degree  it  is  only 
necessary  to  find  two  points  on  the  graph  and  draw  a  straight 
line  through  them. 

E.g.  In  graphing  the  equation  a:  —  y  =  5,  we  choose  x  =  0  and  find 
y  =  —  5,  and  choose  y  —  0  and  find  x  =  b  and  plot  the  points  (0,  —  5) 
and  (5,  0).     The  line  through  these  points  is  the  one  required. 


112 


GRAPHIC  REPRESENTATION 


EXERCISES 

Construct  the  graph  for  each  of  the  following  equations.  From 
each  graph  tell  whether  y  increases  or  decreases  as  x  increases. 

1.  3a;-f-2?/  =  l.         5.    5  —  2y  =  ^x.  9.   3a;  — 42/=— 7. 

2.  5a;-32/=-3.     6.    3x  +  6y'=-15.     lo.   2>x-4.y=-12. 


3.    lx  +  10y  =  2.       7. 


2/  =  0. 


11.   72/  =  9a;-63. 
4.    a; 4-22/  =  0.  8.    3a;  — 4 2/ =  7.  12.   a;  =  52/ +  3. 

116.   Illustrative  Problem.    Graph  on  the  same  axes  the  two 
equations  aj4-2/  =  4  and  y  —  x  =  2. 


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Solution.  The  two  graphs  are  found  to  intersect  in  the  point  (1,  3). 
Since  the  point  lies  on  both  lines,  its  coordinates  should  satisfy  both 
equations,  as  indeed  they  do.  Since  these  lines  have  only  one  point 
in  common,  there  is  no  other  pair  of  numbers  which,  when  substituted 
for  the  variables  x  and  y,  can  satisfy  both  equations. 


GRAPHIC  REPRESENTATION  OF  EQUATIONS 


113 


Hence,  x  =  l,  y  =  3,  which  is  written  (1,  3),  is  called  the 
solution  of  this  pair  of  equations. 

117.  Definition.  These  two  equations  are  called  independent 
because  their  graphs  are  distinct.  They  are  called  simulta- 
neous because  there  is  at  least  one  pair  of  values  of  x  and  y 
which  satisfy  both. 

118.  Since  two  straight  lines  intersect  in  but  one  point,  it 
follows  that  two  linear  equations  which  are  independent  and 
simultaneous  have  one  and  only  one  solution.  The  solving  of 
two  simultaneous  equations  may  be  regarded  as  finding  the 
coordinates  of  the  point  where  their  graphs  meet.  This  may 
be  done  by  constructing  the  graphs  or  by  other  methods  of 
solution  which  are  considered  in  the  next  chapter. 


EXERCISES 

Graph  the  following  and  thus  find  the  solution  of  each  pair 
of  equations : 


1. 


2. 


5. 


|'2a;-32/  =  25, 
x-\-y  =  ^. 

'  5  a;  +  6  2/  =  7, 
2a5  — ?/=  —  4. 

r5aj  +  32/  =  8, 
\2x-y=-10. 

r6a;  +  8?/  =  16, 
{2x-Zy  =  n. 
r3a:-42/=:l, 

l2a;-72/  =  5. 


6. 


9. 


10. 


(y  +  Zx=l, 
\2y^x^-Q. 

(x-2y  =  2, 
{2x-y=:-2. 
(5x-7y  =  21, 


X—  4  y  =  —  1. 


[5a;  +  22/  =  8, 
2a;-32/=-12. 

-6, 
2/= -12. 


l2x-42/=- 


119.  Not  all  pairs  of  equations  in  two  unknowns  are  inde- 
pendent. 

E.g.  If  we  attempt  to  plot  x  +  p  =  1  and  2  x+2y=2  the  graphs 
will  be  found  to  coincide.  Such  equations  are  called  dependent,  since 
one  can  be  derived  from  the  other. 


114  GRAPHIC   REPRESENTATION 

Not  all  pairs  of  equations  are  simultaneous. 

E.g.  If  we  attempt  to  plot  x  +  y  =  1  and  a;  +  ?/  =  2  the  graphs  will 
be  found  to  be  parallel  and  hence  they  have  no  point  in  common. 
Such  equations  are  called  contradictory,  since  it  is  impossible  for  x -^  y 
to  equal  both  1  and  2  at  the  same  time. 

Historical  Note.  The  representation  of  equations  by  means  of  lines  is 
due  to  Ren6  Descartes  (1596-1650).  (See  also  page  50.)  This  must  be 
regarded  as  one  of  the  greatest  contributions  of  all  time  to  mathematics. 
Not  only  is  it  possible  to  represent  straight  lines  by  equations  as  we  have 
here  but  a  very  large  number  of  curves  of  different  kinds  may  be  so 
represented.  The  points  where  two  curves  meet  are  thus  found  by  solv- 
ing the  equations  which  represent  them.  This  enables  us  to  use  the  opera- 
tions of  algebra  in  solving  a  large  range  of  problems  pertaining  to  lines 

and  curves. 

REVIEW  QUESTIONS 

1.  How  may  a  point  in  a  plane  be  located  by  reference  to 
two  fixed  lines  ?  What  are  these  fixed  lines  called  ?  What 
names  are  given  to  the  distances  from  the  point  to  the  fixed 
lines  ?  Why  are  negative  numbers  needed  in  order  to  locate 
all  points  in  this  manner  ? 

2.  Draw  a  pair  of  axes  in  a  plane  and  locate  the  following 
points :  (5,  0),  (-  2,  0),  (0,  3),  (0,  - 1),  (0,  0). 

3.  How  many  pairs  of  numbers  can  be  found  which  satisfy 
the  equation  x—2y  =  Q>?  State  five  such  pairs  and  plot  the 
corresponding  points.  How  are  these  points  situated  with 
respect  to  each  other  ?  What  can  you  say  of  all  points  corre- 
sponding to  pairs  of  numbers  which  satisfy  this  equation  ? 
What  is  meant  by  the  graph  of  an  equation  ? 

4.  How  many  pairs  of  numbers  will  simultaneously  satisfy 
the  two  equations  Sx-{-2y=  7  and  x-{-y  =  S?  Show  by 
means  of  a  graph  that  your  answer  is  correct. 

5.  Is  the  graph  representing  the  equation  x  —  y  =  S  limited 
in  either  direction?  If  negative  numbers  could  not  be  used, 
would  the  graph  of  this  equation  be  limited  in  either  disection  ? 

6.  If  negative  numbers  could  not  be  used,  how  would  the 
graph  of  ic  -j-  2/  =  3  be  limited  ? 


DRILL   EXERCISES  115 

DRILL  EXERCISES 

1.  From  41  a^+  7  ab  —  oc^—9  ah  subtract  8  a6+  7  c^+SO  a^. 

2.  From  15-\-2x—9  xy  —3y  subtract  5xy  —  4:X  —  2y. 

3.  Subtract  12 -\-S  x— 4:  a  — 6  c  — US abc  from  5  cc— 2  a+3  c. 

4.  2-(nx-{-A2y-Wx-64)-(5-91x-2y-13xy). 

5.  {Sl-2x-\-Sy-5)(x  +  y). 

6.  8a;  +  (134-18a;-6)-(5-6ic)=16a;  +  10. 

7.  5-(7-4a;  +  2  2/-4  6)-(8a;-6  2/-9+  26)4-8a. 

8.  8a;-(-3-2-4-7)4-5x4-(2+6+4)-(-3a;+2). 

9.  5y-\-2x-(6-4:X-5x)-3y-(4:X-2y)-(^7y  +  S). 

10.  35  2/-(41a;-16-12  2/)+5a;  +  (-6+46y-18a;). 

11.  (x-iy-(x-S){2x-l)  =  -x'-{-9S. 

12.  (7  +  a;)(a;-4)  +  (l-fl;)2=-23+2a;2. 

13.  (12-4a;)(2-a;)-4(l  +  a;)2  =  5a;4-119. 

14.  (a;  - 17)  (59  -  2  a;)  -  (1 -a;)2= (6- 3  a;)(a;- 2)  4- 384.     . 

15.  (3x-2)^(x-iy-\-(x-2y=2(x-l)(x-2)-^5, 

16.  (6  -3x)(2  +  a;)+16(aj-l)2=13(a;+4)2+364. 

17.  -1.+     ^  ' 


18. 


x  —  b     X— a     X  —  c 
7  3  4 


x  —  a     x-\-a     X  -{-c 

19.  Solve  for  each  letter  in  terms  of  the  others : 

a     b     c 

20.  The  sum  of  two  numbers  is  s  and  one  of  them  is  x. 
How  do  you  represent  the  other  ? 

21.  The  difference  between  two  numbers  is  d  and  one  of 
them  is  x.     How  do  you  represent  the  other  ? 

22.  How  do  you  represent  the  area  of  a  rectangle  whose 
width  and  length  are  w  and  I  feet  respectively  ?  How  do  you 
represent  the  area  of  a  rectangle  which  is  a  feet  wider  and  b 
feet  longer  than  this  one  ? 


CHAPTER   VIII 

SIMULTANEOUS   LINEAR   EQUATIONS 

120.  Simultaneous  linear  equations  have  already  been  con- 
sidered in  connection  with  graphs.  In  this  chapter  such 
equations  will  be  solved  by  the  algebraic  method  called  elimi- 
nation. The  process  consists  in  combining  the  equations  in 
such  a  way  as  to  get  rid  of  one  of  the  unknown  numbers. 

ELIMINATION  BY  ADDITION  OR  SUBTRACTION 

121.  Illustrative  Examples.     Solve 

a; +  2  2/ =  7,  (1) 

\3a;-2y  =  5.  (2) 

Adding  the  members  of  these  equations,  +  2  y  and  —  2  y  cancel. 
Hence,  4  a;  =  12,  and  x  —  Z. 

Substituting  in  (1),         3  +  2  ?/  =  7,  and  y  —  2. 
Verify  this  by  drawing  the  graphs,  and  also  by  substituting  a;  =  3, 
y  =  2,  in  (1)  and  (2). 

If  one  of  the  variables  does  not  already  have  the  same  co- 
efficient in  both  equations,  we  proceed  as  follows : 

Solve  the  equations 

7aj  +  32/  =  4-2/  +  4a;,  (1) 

3aj-2;  =  42/-2-a;.  (2) 

From  (1)  by  ^,  5,  3  a:  +  4  ?/  =  4.  (3) 

From  (2)  by  ^,  5,  4  a;  -  5  ?/  =  -  2.  (4) 

(3)  .  4  12  a:  +  16  y  =  16.  (5) 

(4)  .  3  12  a?  -  15  ?/  =  -  6.  (6) 
(5) -(6)  31 2/ =  22.  (7) 
2)131  2/  =  If.  (8) 
Substituting  in  (3),  x  =  ^f.  (9) 
Hence,  the  solution  is  ^f,  |f. 

116 


ELIMINATION   BY   ADDITION   OR   SUBTRACTION       117 

(3)  •  4  means  that  the  members  of  equation  (3)  were  multiplied  by 
4,  (5)  _  (6)  means  that  the  members  of  equation  (6)  were  subtracted 
from  those  of  (5). 

The  process  used  in  the  solution  just  given  is  called  elimina- 
tion by  addition  or  subtraction. 

Make  a  rule  for  solviug  a  pair  of  equations  by  this  process. 

EXERCISES 

Solve  the  following  pairs  of  equations  by  addition  or  sub- 
traction. Substitute  the  results  in  the  given  equations  in  each 
case  to  test  the  accuracy  of  the  solution. 

2  a;  +  3  y  =  22,  f  7  m  =  2  7i  -  3, 


(2x-\-Sy  =  22,  ^^     I 

\x  —  y  =  l.  '    \ 

I  cc  —  y  =  6.     .  I 


19n  =  6m  +  89. 
15  1c  =  10-  20  ?, 
25A:-30Z  =  80. 
'6x-{-30  =  8  7j,  r6c-hl5cZ=:-6, 

,Sy-\-lT  =  2-3x.  ^^'    l21d-8c  =  -74. 

(Sx-4:y  =  12x,  (2x-3y  =  4:, 

[4:X-^2y=z3-\-4:y.  '    [2y-3x  =  -21. 

=  27, 
19  - 1  w. 


(x  +  6y=:2x-16,  (u  +  v 

'    l3a;-22^  =  24.  *    11^  = 

(5x  +  10y  =  ^T,  r7a  =  l  +  102/, 

*    l2a;  +  5y  =  -2.  •'    \l6jy  =  10a-l. 

r  5  a; +  3  2/  = -2,  f  28  a; -f- 14  ?/ =  23, 

'^'    [3x-^2y  =  -l.  ^'^'    [Ux^Uy  =  l. 

(3a  +  7b  =  7y  (5x-^2y  =  x-\-18 

l5a  +  36  =  29.  ^^'    \2x-\-3y  =  3x-^'. 


rr  =  3s-19,  r7y-a;  =  a;-J 

9      J  '  19      1 

•     U  =  3r-23.  '    {2y-^3x  =  3S. 

r  2^9  =  5^-16,  (6x  +  2y  =  -2, 

^'     [7q  =  -3p-{-5.  ^^'     U-42/  =  -35. 


118 


SIMULTANEOUS   LINEAR   EQUATIONS 


ELIMINATION  BY  SUBSTITUTION 

122.   Illustrative  Problem.     Solve  the  equations : 

.5x  —  6y  =  —  S. 

13 -2x 


From  (1)  by  6f  and  A 

Substituting  in  (2), 

By  F,  V, 

By  IV,  VII, 

By  I,  A, 

By  A 

From  (3)' 

y  = 


3 


^^_6(13-2.) 


8. 


5x-2(13-2j:)  =  -8. 
5a:-26  +  4a:  =  -8. 
9  a;  =  18. 

x  =  2. 

13-2-2      o 
V  = =  3. 


(1) 

(2) 

(3) 

(^) 

(5) 
(6) 
(7) 
(8) 

(9) 


Verify  this  by  substituting  these  values  of  x  and  y  in  (1)  and  (2). 

The  process  here  used  is  called  elimination  by  substitution. 

Make  a  rule  for  solving  a  pair  of  equations  by  this  process. 

This  process  is  usually  convenient  when  one  of  the  variables 
has  a  coefficient  unity  in  one  equation  and  not  in  the  others, 
since  in  that  case  no  fractions  are  introduced  by  the  substitution. 


EXERCISES 


1. 


2. 


5. 


r  a;  -+-  2  2/  =  4, 
[2x  +  y  =  5, 

[3x-y  =  5^ 
[5x-\-2y  =  23. 

(2x-^y  =  S, 
[Sx-7y  =  S0. 

'  5  2/  +  a;  =  7, 
5x-'3y  =  4:-2x-[-7. 

'5x-{-Sy  =  -l, 
.6y  —  x  =  4:y  —  7. 


6. 


7. 


8. 


9. 


10. 


x-y=^S7, 

2x-^Sy  =  Slx-i-lSy. 

2x  —  y  =  y  +  6, 
x  +  2y  =  4.y-\-3. 


(5x-\- 
l2a;  + 


3  2/  =  0, 

2/  =  l. 


3x-2y  =  3, 
2x-{-3y  =  6x- 

(5x-3y  =  0, 
[2x  —  6y  =  —  x, 


SOLUTION   BY   ELIMINATION 


119 


Solve  by  either  method 
[x-y=10. 

7x-5y='S3. 


EXERCISES 

of  elimination : 


3. 


4. 


6. 


7. 


10. 


11. 


12. 


(Sx-4y  =  S, 
[ix-^3y  =  -6. 

(2x-4y  =  S, 
l3x-\-2y  =  4. 

(3x-4.y  =  8, 
\2x-j-3y  =  n. 


f4.y-2x  =  3, 
[2y-^5x==6. 

g      (3x-y=2x-l, 
\l2x  +  y  =  U. 

9      (2x  +  3y  =  5, 
1  6  a;  + 14  2/  =  0. 


|4a;  +  3y  =  5, 
l7a;-22/  =  74. 

r6y  +  2a;  =  ll, 
1 3  2/ +  12  a;  =  18. 

Uy  +  9x=-5, 
L  a?  +  2/  =  —  5. 


13. 


14. 


15. 


16. 


(3y-\-5x  =  12-\-2x, 
1 17  a;  —  2/  =  4  2/  —  20. 
r6  +  a;  +  2/  =  2a;-l, 
l32/  +  .T=6y  +  9. 
r2/  +  5a;  =  2a;4-5, 
l22/-3a;  =  19. 

r6a;  +  22/=23, 
llOa;-52/  =  21. 


18. 


19. 


20. 


Ans.x  =  3^,  y  =  2^\. 

17.     r3a.-72/  =  15, 
1  5  a;  +  4  2/  =  11. 

^ns.  a;  =  -Vf,  2/  =  -J^. 
7a;-42/  =  3, 
[5x-\-Sy=6. 

'12y-10x  =  -6, 
.7y  +  a;  =  99. 

7a;-32/  =  -7, 
52/-9a;  =  l. 


21. 


22. 


23. 


24. 


( 

(7x-\-4.y  =  3, 
\2x-\-3y  =  25. 
r34a;  +  702^  =  4, 
l5a;~82^  =  -36. 

(7x  +  9y  =  S, 
\2x^3y  =  21, 
8a;  4-4^  =  49, 
5a;-8y  =  28, 


120 


SIMULTANEOUS  LINEAR  EQUATIONS 


123.  The  equations  thus  far  given  have  for  the  most  part 
been  written  in  a  standard  form,  ax -\- by  =  c,  in  which  all  the 
terms  containing  x  are  collected,  likewise  those  containing  y, 
and  those  which  contain  neither  variable.  When  the  equations 
are  not  given  in  this  form,  they  should  be  so  reduced  at  the 
outset,  as  in  the  following. 


5              2          ^' 

(1) 

Example:   Solve 

5x-2     22/  +  1     o 
[      3       '        5 

(2) 

Solution.     (1)  .  10, 

14  2^  -  8  +  10  X  -  15  =  15. 

(3) 

Transposing  in  (3) 

Hy  +  10x  =  38. 

W 

(4)^2, 

7y  +  5a;  =  19. 

(5) 

(2)  .  15, 

25  x  -  10  +  6  1/  +  3  =  30. 

(6) 

Transposing  in  (6), 

25  a:  +  6  y  =  37. 

(7) 

(5)  .  5, 

25  a; +  35!/ =  95. 

(8) 

(8)  -  (7), 

29  y  =  58. 

(9) 

(9)  s-  29, 

2/ =  2. 

(10) 

Substituting  in  (7), 

x=l. 

Check.  Substitute  x  =  1,  y  =  2  in  (1)  and  (2)  and  see  that  each  is 
satisfied. 

EXERCISES 

After  reducing  each  of  the  following  pairs  of  equations  to 
the  standard  form,  solve  by  means  of  either  process  of  elimi- 
nation : 


^      {x-U  =  7y, 
I  6  2/  -f  1  =  07. 


4. 


2. 


'■( 


16  ic  —  3  ?/  =  7  a:, 
42/  =  7aj+-5. 

r  +  l  =  -4s, 
25  =  13-5r. 


r       i-8n 

m  =  — - — , 
5 

3m  +  5^1 


1. 


( m  — 

13  7M 


n  =  16, 
=  8-2». 


SOLUTION  BY  ELIMINATION 


121 


6. 


7X-15 


=  2/, 


2  x  —  y  =  3. 
(x-S 


14 


.  , 


8a-3  .  56-2_.o 
2a+7     36  +  10 


10 


-3f 


7.    ■{     5y 


[x-^7y  =  6. 

I  3a;-5  =  -y, 
[8y-h76  =  5x 

ra  +  46  =  14, 
1 3  a- 6  =  14. 


15. 


r7v-4  ,  2x-S_r, 

I  6a;-3      2y-\-l^rr 

15  5 

Ans.  rc  =  5^g-,  i/  =  2||. 


^ns.  a  =  53%,  b  =  2j%. 


(x+y     x—y_ 
10.    I     2     "^     2 
[2a;-y  =  16. 

'  x  —  y  ,x-\-y 


+  r_^  =  10,  16. 


3i/  +  7     5a;-7 


=  10, 


11.    \ 


+ 


=  8, 


2  a;— y      3j/--^_^oi 


17. 


12. 


13. 


2  4 

[7m-h8     7n  — 1 

5  4 

2m  — 4  ,  71  —  1 


2               3 
2^-4     2y-l  oi 

-^ 4~— ^^• 


'5+3p_5£-2^_2 
7  4 

6^)4-8^  =  108. 


-2, 


+  ■ 


=  -i- 


18. 


x-S     y-\-S>      ^ 


3a;-22/  =  4, 
2x-l_7j-4__^g 
5  3 


5aj  +  72/  =  89J, 


a;+7      2y-4  19.       2a:-4      Gy-l^-^g^ 

2     ^      7      ~  .3^5  ^ 


r32a;-92/  =  299, 
20.    ]2r»-5      3?/-l 


=  -16. 


122  SIMULTANEOUS  LINEAR  EQUATIONS 

SIMULTANEOUS  LITERAL  EQUATIONS 
Examples,     l.    The  sum  of  two  numbers  is  35  and  their  dif- 
ference is  5.     What  are  the  numbers  ? 

Let  X  represent  one  number  and  y  the  other.     Form  two  equations 
and  solve  them. 

2.  The  sum  of  two  numbers  is  48  and  their  difference  is  24. 
What  are  the  numbers  ? 

3.  The  sum  of  two  numbers  is  411  and  their  difference  is 
23^.     What  are  the  numbers  ? 

4.  The  sum  of  two  numbers  is  8590  and  their  difference  is 
3480.     What  are  the  numbers  ? 

5.  If  the  sum  of  two  numbers  is  s  and  their  difference  is  c?, 
find  the  number. 

Solution.     Let  x  represent  one  of  the  numbers  and  y  the  other. 
Then,  x  +  y  =  s,  (1) 

x-y  =  d.  (2) 

Solving,  we  get  :,==i±i:.^+| 

-d      s     d 


and  y  = 

_  2         2     2 

Translated  into  words  these  results  are : 

Given  the  sum  of  any  two  numbers  and  their  difference.  Then 
one  of  the  numbers  is  half  their  sum  plus  half  their  difference  and 
the  other  is  half  their  sum  minus  half  their  difference. 

6.   Test  this  general  solution  by  applying  to  tlie  following : 

s  =  48,      d  =  24.  5  =  40,  d  =  52. 

5  =  8590,  d  =  348.  s  =  38,  d  =  50. 

In  the  following  x  and  y  are  the  unknowns.     Solve  for  them 

in  terms  of  the  other  letters. 

mx  -\-2ny  =  k, 


\    ax  -\-  by  =  1,  ■     ^-    I  Q  I  o  1 

7     J         ^     -^        '  [S-{-2mx=  -{-ny. 


cx—dy  =  1. 

c-d,  10. 


[2x-Sy 
'    Ux-2y 


8.     .  „         ^ 

c  +  d. 


x  +  y  =  a, 
^  y     -1. 


a+b      a—b 


DRILL   EXERCISES 


123 


DRILL  EXERCISES 

1.  {x-2y-(x-l)(x  +  2)  =  6-5x, 

2.  (x-2)(x+2)-{-{3x-l)(2-x)  =  (x-2X5-2x), 

3.  (x-Sy-\-(2x-{-5y=(5x-S)(x-{-5)-7. 

4.  (2-xy-(2x-iy=(-Sx  +  l)(4.  +  x)-4:, 

(2x-3y-l  =  0, 
\5x-i-2y  =  12. 

In  the  following  x  and  y  are  the  unknowns.    Solve  for  them 
in  terras  of  the  other  letters. 


6. 


7. 


x-\-y  =  a, 

x-y  =  b.  8. 

'  ax-]-by  =  c, 
ax  —  by  =  d. 
9.    (x-ay  +  (x-hy=2(x-cy. 
10.    ax  —  a(h  —  a;)  -f-  ac  =  3  ah. 
2 x—a     2 x—b 


X  ,  y 
a     b 

a     0 


11 


=  1. 


12. 


x  —  a 


x-\-b       2x-\-a  x-\-a     x—h 

13.  2ax  —  4:by-\-2cx—{cx  —  Zax  —  i:by). 

14.  Find  the  value  of  a^-3ca;H-8c  for  a;  =  l,  a  =  2,  6  =  3, 
c  =  4.  &  +  ab-W 

15.  A  father  is  now  twice  the  age  of  his  son.  If  x  represents 
the  son's  age  now,  express  twice  the  sum  of  their  ages  5  years  ago. 

16.  One  number  is  3  less  than  4  times  another  number.  If 
X  represents  one  of  the  numbers  express  one  third  the  sum  of 
the  numbers. 

17.  If  h  is  the  digit  in  hundreds'  place,  t  the  digit  in  tens' 
place,  and  u  the  digit  in  units'  place  of  a  certain  number, 
express  the  number  obtained  by  inverting  the  order  of  the 
digits  of  the  given  number. 

18.  /i,  t,  and  u  are  the  digits  in  hundreds',  tens',  and  units' 
places  of  a  number.  Express  the  number  obtained  by  increas- 
ing each  digit  by  2. 


124  SIMULTANEOUS   LINEAR   EQUATIONS 

124.  Many  problems  may  be  solved,  using  either  one  or  two 
unknowns. 

Ex.  Find  two  numbers,  whose  sum  is  20,  such  that  when  one 
of  them  is  subtracted  from  twice  the  other,  the  remainder  is  16. 

(a)  Using  one  unknown.  Let  x  represent  one  number.  Then  20  — a; 
is  the  other  number,  and  the  equation  is  2  a;  —  (20  —  x)  =  16. 

(6)  Using  two  unknowns.  Let  x  and  y  represent  the  two  numbers. 
Then,  (    -  +  2/  =  20, 


2x-y=lQ. 

The  translation  of  problems  into  equations  is  usually  easier 
when  more  than  one  unknown  is  permitted.  This  is  due  to 
the  fact  that  in  this  case  each  of  the  given  relations  between 
'the  numbers  is  put  down  as  a  separate  equation. 

PROBLEMS    INVOLVING  TWO  VARIABLES 

1.  li  w  and  I  are  the  width  and  length  of  a  rectangle, 
express  its  area  and  also  its  perimeter  in  terms  of  w  and  I. 

2.  If  the  width  of  the  rectangle  in  the  preceding  is  increased 
by  10  and  its  length  by  20,  express  its  new  perimeter  and  also 
its  new  area  in  terms  of  w  and  I. 

3.  If  X  and  y  represent  the  ages  of  a  father  and  son  respec- 
tively, represent  the  sum  of  their  ages  5  years  ago  in  terms  of 
X  and  y ;  also  the  difference  of  their  ages  8  years  hence. 

4.  If  a  number  consisting  of  two  digits  is  increased  by  15 
by  changing  the  order  of  its  digits,  which  is  greater,  the  digit 
in  tens'  or  in  units'  place  ? 

5.  If  a  number  is  decreased  by  changing  the  order  of  its 
digits,  which  is  greater,  the  digit  in  tens'  or  in  units'  place  ? 

In  solving  the  following  problems,  use  two  variajples  in  each 
case: 

6.  A  rectangular  field  is  32  rods  longer  than  it  is  wide. 
The  length  of  the  fence  around  it  is  308  rods.  Find  the  di- 
mensions of  the  field. 


PROBLEMS   INVOLVING  TWO   VARLiBLES  125 

7.  Find  two  numbers  such  that  7  times  the  first  plus  4 
times  the  second  equals  37;  while  3  times  the  first  plus  9 
times  the  second  equals  45. 

8.  A  certain  sum  of  money  was  invested  at  5  %  interest 
and  another  sum  at  6%,  the  two  investments  yielding  $980 
per  annum.  If  the  first  sum  had  been  invested  at  6%  and 
the  second  at  5  %,  the  annual  income  would  be  $1000.  Find 
each  sum  invested. 

9.  The  combined  distance  from  the  sun  to  Jupiter  and 
from  the  sun  to  Saturn  is  1369  million  miles.  Saturn  is  403 
million  miles  farther  from  the  sun  than  Jupiter.  Find  the 
distance  from  the  sun  to  each  planet. 

10.  Find  two  numbers  such  that  7  times  the  first  plus  9 
times  the  second  equals  116,  and  8  times  the  first  minus  4 
times  the  second  equals  4.  «. 

11.  The  sum  of  two  numbers  is  108.  8  times  one  of  the  num- 
bers is  9  greater  than  the  other  number.     Find  the  numbers. 

12.  Two  investments  of  $24,000  and  $16,000  respectively 
yield  a  combined  income  of  $840.  The  rate  of  interest  on 
the  larger  investment  is  1  %  greater  than  that  on  the  other. 
Find  the  two  rates  of  interest. 

13.  A  father  is  twice  as  old  as  his  son.  Twenty  years  ago  the 
father  was  six  times  as  old  as  his  son.     How  old  is  each  now  ? 

14.  If  the  length  of  a  rectangle  is  increased  by  3  feet  and 
its  width  decreased  by  1  foot,  its  area  is  increased  by  3  square 
feet.  If  the  length  is  increased  by  4  feet  and  the  width  de- 
creased by  2  feet,  the  area  is  decreased  by  3  square  feet. 
What  are  the  dimensions  of  the  rectangle  ? 

Note  that  if  w  and  I  are  the  original  width  and  length  of  the  reo* 
tangle,  the  term  Iw  will  cancel  out  of  both  equations. 

,  15.  A  steamer  on  the  Mississippi  makes  6  miles  per  hour 
going  against  the  current  and  19|-  miles  per  hour  going  with 
the  current.  What  is  the  rate  of  the  current  and  at  what  rate 
can  the  steamer  go  in  still  water  ? 


126  SIMULTANEOUS    LINEAR   EQUATIONS 

16.  A  starts  at  7  a.m.  for  a  walk  in  the  country.  At  10  a.m. 
B  starts  on  horseback  to  overtake  A,  which  he  does  at  1  p.m. 
If  the  rate  of  B  had  been  two  miles  per  hour  less,  he  would 
have  overtaken  ^  at  4  p.m.     At  what  rate  does  each  travel  ? 

17.  A  camping  party  sends  a  messenger  with  mail  to  the 
nearest  post  office  at  5  a.m.  At  8  a.m.  another  messenger  is 
sent  out  to  overtake  the  first,  which  he  does  in  2J  hours.  If 
the  second  messenger  travels  5  miles  per  hour  faster  than  the 
first,  what  is  the  rate  of  each  ? 

18.  There  are  two  numbers  such  that  3  times  the  greater  is  18 
times  their  difference,  and  4  times  the  smaller  is  4  less  than 
twice  the  sum  of  the  two.     What  are  the  numbers  ? 

19.  A  picture  is  3  inches  longer  than  it  is  wide.  The  frame  4 
inches  wide  has  an  area  of  360  square  inches.  What  are  the 
dimension*  of  the  picture  ? 

20.  The  difference  between  2  sides  of  a  rectangular  wheat 
field  is  30  rods.  A  farmer  cuts  a  strip  5  rods  wide  around  the 
field,  and  finds  the  area  of  this  strip  to  be  7^  acres.  What 
are  the  dimensions  of  the  field  ? 

21.  The  sum  of  the  length  and  width  of  a  certain  field  is 
260  rods.  If  20  rods  are  added  to  the  length  and  10  rods 
to  the  width,  the  area  will  be  increased  by  3800  square  rods. 
What  are  the  dimensions  of  the  field  ? 

22.  In  a  number  consisting  of  two  digits  the  sum  of  the 
digits  is  12.  If  the  order  of  the  digits  is  reversed,  the  number 
is  increased  by  36.     What  is  the  number  ? 

23.  A  bird  attempting  to  fly  against  the  wind  is  blown  back- 
ward at  the  rate  of  7^  miles  per  hour.  Flying  with  a  wind  ^  as 
strong,  the  bird  makes  48  miles  an  hour.  Find  the  rate  of  the 
wind  and  the  rate  at  which  the  bird  can  fly  in  calm  v/eather. 

24.  There  is  a  number  whose  two  digits  differ  by  2.     If  the. 
digit  in  units'  place  is  multiplied  by  3  and  the  digit  in  tens' 
place  is  multiplied  by  2,  the  number  is  increased  by  44.     Find 
the  number,  the  tens'  digit  being  the  larger. 


FROBLEMS   INVOLVING   TWO   VARIABLES  127 

25.  In  a  number  consisting  of  two  digits  one  digit  is  equal 
to  twice  their  difference.  If  the  order  of  the  digits  is  reversed, 
the  number  is  increased  by  18.     Find  the  number. 

26.  If  the  length  of  a  rectangle  is  doubled  and  8  inches  added 
to  the  width,  the  area  of  the  resulting  rectangle  is  180  square 
inches  greater  than  twice  the  original  area.  If  the  length  and 
width  of  the  rectangle  differ  by  10,  what  are  its  dimensions  ? 

27.  There  is  a  number  consisting  of  three  .digits,  those  in  tens' 
and  units'  places  being  the  same.  The  digit  in  hundreds'  place 
is  4  times  that  in  units'  place.  If  the  order  of  the  digits  is  re- 
versed, the  number  is  decreased  by  594.    What  is  the  number  ? 

28.  A  man  rowing  against  a  tidal  current  drifts  back  2^ 
miles  per  hour.  Rowing  with  this  current,  he  can  make  14^ 
miles  per  hour.  How  fast  does  he  row  in  still  water  and  how 
swift  is  the  current  ? 

29.  Flying  against  a  wind  a  bird  makes  28  miles  per  hour, 
and  flying  with  a  wind  whose  velocity  is  2f  times  as  great,  the 
bird  makes  46  miles  per  hour.  What  is  the  velocity  of  the 
wind  and  at  what  rate  does  the  bird  fiy  in  calm  weather? 

30.  A  freight  train  leaves  Chicago  for  St.  Paul  at  11  a.m. 
At  3  and  5  p.m.  respectively  of  the  same  day  two  passenger 
trains  leave  Chicago  over  the  same  road.  The  first  overtakes 
the  freight  at  7  p.m.  the  same  day,  and  the  other,  which  runs 
10  miles  per  hour  slower,  at  3  a.m.  the  next  day.  What  is 
the  speed  of  each  ?    . 

31.  Two  boys,  A  and  B,  trying  to  determine  their  respective 
weights,  find  that  they  balance  on  a  teeter  board  when  ^  is  6 
feet  and  A  5  feet  from  the  fulcrum.  If  B  places  a  30-pound 
weight  on  the  board  beside  him,  they  balance  when  J5  is  4  and 
A  5  feet  from  the  fulcrum.     How  heavy  is  each  boy  ? 

32.  S  10,000  and  $  8000  are  invested  at  different  rates  of 
interest,  yielding  together  an  annual  income  of  $  820.  If  the 
first  investment  were  S  12,000  and  the  second  $  6000,  the  yearly 
income  would  be  $  840.     Find  the  rates  of  interest. 


128  SIMULTANEOUS  LINEAR  EQUATIONS 

SIMULTANEOUS  EQUATIONS  IN  THREE  VARIABLES 

125.  Illustrative  Problem.  Three  men  were  discussing  their 
ages  and  found  that  the  sum  of  their  ages  was  90  years.  If 
the  age  of  the  first  were  doubled  and  that  of  the  second  trebled, 
the  aggregate  of  the  three  ages  would  then  be  170.  If  the 
ages  of  the  second  and  third  were  each  doubled,  the  sum  of 
the  three  would  be  160.     Find  the  age  of  each. 

Solution.  Let  a:,  y^  and  z  represent  the  number  of  years  in  their 
ages  in  the  order  named. 

Then,  x  +  i/-\-z  =  90,  (1) 

2x-\-dy  +  z  =  170,  (2) 

and  x  +  2y  +  2z  =  160.  (3) 

If  we  subtract  (1)  from  (2),  we  obtain  a  new  equation  from  which 
z  is  eliminated. 

I.e.  x  +  2y=:S0.  (4) 

Again,  multiplying  (2)  by  2  and  subtracting  (3), 

3x  +  4:y  =  180.  (5) 

(4)  and  (5)  are  two  equations  in  the  two  variables  x  and  y.  Solving 
these  by  eliminating  y,  we  find        x  =  20.  (6) 

Substituting  a;  =  20  in  (4),  y  =  30.  (7) 

Substituting  x  and  y  in  (1),         z  =  40.  (8) 

Check  by  showing  that  the  values  of  x,  y,  and  z  satisfy  the  original 
equations  and  the  conditions  of  the  problem. 

The  values  of  x,  y,  and  z  as  thus  found  constitute  the  solution 
of  the  given  system  of  equations. 

Evidently  x  could  have  iDeen  eliminated  first,  using  (1),  (2),  and 
(1),  (3),  giving  a  new  set  of  two  equations  in  y  and  z.  Let  the  student 
find  the  solution  in  this  manner. 

Also  find  the  solution  by  first  eliminating  y,  using  (1),  (2), 
and  (2),  (3),  getting  two  equations  in  x  and  2,  from  which  the  values 
of  X  and  z  can  be  found. 

126.  Definition.  An  equation  is  said  to  be  of  the  first  degree 
in  Xj  2/,  and  z  if  it  contains  these  letters  in  such  a  way  that  no  one 
of  them  is  multiplied  by  itself  or  by  one  of  the  others  (§  115). 


SYSTEMS   INVOLVING  THREE   VARIABLES 


129 


The  fact  that  the  solutions  are  found  to  be  the  same,  no 
matter  in  what  order  the  equations  are  combined,  indicates 
that  a  system  of  three  independent  and  simultaneous  equations  of 
the  first  degree  in  three  variables  has  one  and  only  one  solution. 

As  in  the  case  of  two  equations,  each  should  be  first  reduced 
to  a  standard  form  in  which  all  the  terms  containing  a  given 
variable  are  collected  and  united  and  all  fractions  removed. 
Make  a  rule  for  solving  a  system  of  three  equations  each  of  the 
first  degree  in  three  variables. 


EXERCISES 

Solve  the  following  systems,  and  check  the  results; 


1. 


2. 


3. 


5. 


2  «  —  2/ H- 2  =  18, 
x-2y  +  ^z  =  10, 
3a;-+-2/_4;z  =  20. 
5a;  —  32/4-2  =  15, 
a;  +  32/-2!  =  3, 
2x  —  y-\-z  —  ^. 

4a;  +  22/  +  2:  =  13, 
x  —  y-\-z  =  ^y 
x-\-2y  —  z  =  l. 
6a;  +  42/  —  42=—  4, 
4:X-2y-hSz  =  0, 
x  +  y-hz  =  ^. 
x-\-2y-{-3z  =  5j 
4:X  —  3y  —  z=:5, 
x  +  y  -\-z  =  2. 
2x-Sy  +  Sz  =  2, 
x  —  4:y-\-5z==lf 
Sx-10y-z  =  5. 


»  4-  .V  4-  2  =  1, 
7.    {x  +  Sy-{-2z  =  S, 
2x-\-Sy-Sz  =  15. 
2x-3y-i-z  =  5, 
3x-\-2y  —  zz=5, 
x  +  y  +  z  =  3, 
v  +  y-\-z  =  6, 
3x-2y-z  =  13, 
2x-y  +  3z  =  26. 
x-{-y-j-z  =  ej 

10.  {Ax  — y—z=  —ly 
2x-\-y-3z==-6. 
2x-3y-4:Z  =  17, 

11.  ■j4a;  — 4?/4-22  =  — 10, 
7x  +  7y  +  5z  =  17. 

{x-\-y-\-z  =  0, 
5  a;  4- 3  2/4- 4  2;  =  — 1, 
2x-7y-^6z  =  21. 


130  SIMULTANEOUS   LINEAR   EQUATIONS 

PROBLEMS  INVOLVING  THREE  VARIABLES 

127.  Illustrative  Problem.  A  broker  invested  a  total  of 
$  15,000  in  the  street  railway  bonds  of  three  cities,  the  first 
investment  yielding  3%,  the  second  3|-%,  and  the  third  4%, 
thus  securing  an  income  of  $535  per  year.  If  the  second 
investment  was  one-half  the  sum  of  the  other  two,  what  was 
the  amount  of  each  ? 

Solution.  Suppose  x  dollars  were  invested  at  3  %,  y  dollars  at  3^%, 
and  z  dollars  at  4%. 

Then,  x  +  y  +  z=  15000,  (1) 

.03  X  +  .035  y  +  Mz  =  535,  (2) 

and  x-\-z  =  2y.  (3)* 

From  (3),  x-~2y  -\-z  =  0.  (4) 

Subtract  (4)  from  (1),  2y  =  15000,  (5) 

and  y  =  5000.  (6) 

From  (1),  by  M,       .035  x  +  .035  y  +  .035  z  =  525.  (7) 

Subtract  (7)  from  (2),        -  .005  x  +  .005  z  =  10.  (8) 

Divide  (8)  by  .005,  -x  +  z  =  2000.  (9) 

Substitute  (6)  in  (4),  x  +  z  =  10000.  (10) 

Add  (9)  and  (10),  2z  =  12000.  (11) 

z  =  6000.  (12) 

Substitute  (6)  and  (l2)  in  (1),  x  =  4000.  (13) 

Hence,  $4000,  $5000,  and  $6000  were  the  sums  invested. 

Solve  the  following  problems,  using  three  unknowns : 

1.  The  sum  of  three  angles.  A,  B,  and  C  of  a  triangle  is 
180  degrees,  i  of  ^  4-  i  of  J5  +  ^  of  0  is  48  degrees,  while  ^ 
of  A-\-\  oi  J5  4- 1  of  C  is  30  degrees.  How  many  degrees  in 
each  angle  ? 

2.  The  combined  weight  of  1  cubic  foot  each  of  compact 
limestone,  granite,  and  marble  is  535  pounds.  1  cubic  foot  of 
limestone,  2  of  granite,  and  3  of  marble  weigh  together  1041 
pounds,  while  1  cubic  foot  of  limestone  and  1  of  granite  to- 
gether weigh  195  pounds  more  than  1  cubic  foot  of  inarble. 
Find  the  weight  per  cubic  foot  of  each  kind  of  stone. 


PROBLEMS   INVOLVING  THREE   VARIABLES  131 

3.  A  number  is  composed  of  3  digits  whose  sum  is  7.  If 
the  digits  in  tens'  and  hundreds'  places  are  interchanged, 
the  number  is  increased  by  180;  and  if  the  order  of  the 
digits  is  reversed,  the  number  is  decreased  by  99.  What  is 
the  number  ? 

4.  The  sum  of  the  angles  A,  B,  and  (7  of  a  triangle  is  180 
degrees.  If  B  is  subtracted  from  C,  the  remainder  is  ^  of  A, 
and  when  C  is  subtracted  from  twice  A,  the  remainder  is  4 
times  B.     How  many  degrees  in  each  angle? 

5.  The  sum  of  the  three  sides  a,  6,  c  of  a  certain  triangle  is 
35,  and  twice  a  is  5  less  than  the  sum  of  b  and  c,  and  twice  c 
is  4  more  than  the  sum  of  a  and  b.  What  is  the  length  of 
each  side  ? 

6.  If  X  is  the  number  of  seconds  in  the  Eastern  inter- 
collegiate record  for  a  mile  run,  y  the  number  in  the  Western 
intercollegiate  record,  and  z  the  number  in  the  world's  record, 
then 

a;4-y  + 2  =  768.97, 

—  x  +  2'y  +  z  =  518.95, 

2a;-^  +  2  =  502.75. 

7.  If  X  is  the  number  of  seconds  in  the  Eastern  inter- 
collegiate record  for  a  half  mile  run,  y  the  number  in  the 
Western  intercollegiate  record,  and  z  the  number  in  the 
world's  record,  then 

2x-\-Sy-{-z  =  692.9, 

3a;  4-2^  +  22=804.6, 

2a;-!/  +  2  =  226.5. 

8.  If  a;  =  number  of  seconds  in  the  world's  mile  trotting 
record  in  1806,  y  =  number  of  seconds  in  the  world's  record  in 


132  SIMULTANEOUS  LINEAR  EQUATIONS 

1885,  and  z  =  number  of  seconds  in  the  world's  record  in  1911, 

then 

X  +  y  4- :3  =  426.25, 

2a;-}-42/  +  62  =  1584, 

-a;  +  2/  +  22  =  186.75. 

9.  Diophantus  of  Alexandria  (see  page  19)  gives  the  follow- 
ing problem  :  ''  Find  three  numbers  such  that  the  sum  of  each 
pair  is  a  given  number.''  Solve  this  problem  if  the  given 
numbers  are  20,  30,  40  (the  numbers  actually  given  by  Dio- 
phantus). 

Diophantus  remarks  that  half  the  sum  of  the  three  given  numbers 
must  be  greater  than  each  singly.  This  is  to  prevent  negative  num- 
bers in  the  results. 

10.  Solve  the  preceding  problem  when  the  given  numbers 
are  a,  h,  c.     That  is,  solve  the  system  x-^y  =  a, 

y  +  z  =  h, 
z-\-x  =  c. 

It  is  interesting  to  note  that  Diophantus  states  his  problem  in 
words  in  its  general  form  but  he  solves  it  for  a  special  case,  viz.  for 
a:  +  2^  =  20,  ?/  +  2  =  30,  2  +  a:  =  40.  The  Greeks  did  not  use  letters  to 
represent  numbers  in  general.  Hence  they  had  no  formulas  such  as 
we  now  have. 

REVIEW  QUESTIONS 

11.  Describe  elimination  by  the  process  of  substitution ;  also 
by  the  process  of  addition  or  subtraction.  Under  what  condi- 
tions is  one  or  the  other  of  these  methods  preferable  ? 

12.  Describe  the  solution  of  a  system  of  three  linear  equa- 
tions in  three  unknowns.  Is  it  immaterial  which  of  the  three 
variables  is  eliminated  first  ? 

13.  Can  you  find  a  definite  solution  for  two  equations  each 
containing  three  unknowns  ?  Illustrate  this  by  means  of  the 
equations  4^x  —  3y  —  z  —  6  and  x-\-y  -{■  z  =  2. 


DRILL   EXERCISES 


133 


DRILL  EXERCISES 


1.  (1-3  xy  +  (2  ic  -f  1)2  =  5  ar'  +  (2  x-\-Q,){4.  x  +  27). 

2.  (l  +  2a;)(2-3a;)-h(a;-4)(a;-h4)=(a;+14)(18-5a;)-l. 

3.  2{x  +  5)(x  -5)  =  (x-  o){x  4- 1)  +  (a^  -  2)^-1. 


4. 


5. 


2 
x-\-l 


"^~3~~    ^' 


7. 


2/ 


=  i- 


a 


2 
x  +  a 


+ 


4 

y—h 
3 

^^^+^  =  1. 


1, 


6. 


a;  +  2/  +  2;  =  6, 
2a;-2/  +  2  =  3, 
3a;  +  22/-22;  =  l. 

iB-2/  +  0  =  4, 
a;  —  22/  +  42;  =  6, 
L2a;  +  2/-32  =  10. 

a;  +  2/  +  2  =  0, 
2a;  —  4?/-f-2=—  3, 
l3a;+2y4-42  =  3. 


3  2 

( 2  (ix  —  Sby  =  Cy  9. 

l2aa;  +  36?/  =  d. 

Solve  the  following  equations  for  x : 

10.  (a;  —  a)(x  —  a)  =  —  a;(a  —  a?)  +  (a  +  6)a;. 

11.  ^=-^. 
6      c  — a? 

4a4a     35a 

3  a;       x       2      ox 

jg     2a-36      b     3a^l3a 

3a;  a;     2a;        6 

14.  Express  the  average  of  the  numbers  3,  8,  —  9,  12 ;  also 
of  the  numbers  3  a,  b,  2  c,  —5  b. 

15.  Express  the  sum  of  the  squares  of  four  consecutive  even 
integers  of  which  2  ri  is  the  smallest. 

16.  Express  the  sum  of  the  squares  of  four  consecutive  odd 
integers  of  which  2  /i  4-  1  is  the  greatest. 

17.  A  wall  is  I  feet  long  and  h  feet  high.  There  are  three 
windows  in  it  7c  feet  wide  and  m  feet  high.  By  how  much 
does  the  area  of  the  wall  exceed  that  of  the  three  windows  ? 
Express  the  result  in  terms  of  I,  h,  k,  and  m. 


PART  TWO 

CHAPTER  IX 

PRODUCTS,  QUOTIENTS,  AND  FACTORS 

The  principles  and  processes  studied  in  Part  One  enabled  us 
to  solve  equations  of  the  first  degree  involving  one  or  more  un- 
knowns, and  to  solve  all  problems  which  we  could  translate  into 
such  equations. 

In  Part  Two  we  begin  with  factoring  which  will  lead  to  the 
solution  of  equations  of  the  second  degree  and  thereby  to  the 
solution  of  problems  still  more  difficult. 

128.  Repeated  Factors.  Number  expressions  containing  re- 
peated factors  have  already  been  considered  in  Chapter  IV. 
X'X  was  written  a^  and  called  tJie  square  of  x,  or  x  square; 
similarly,  (a  -f-  6)  (a  +  b)  was  written  (a  +  by  and  read  the 
square  of  the  binomial  a  +  b  or  a  +  b  squared. 

129.  Definitions.  Any  number  written  over  and  to  the  right 
of  a  number  expression  is  called  an  index  or  exponent.  If  an 
exponent  is  a  positive  integer,  it  shows  how  many  times  the  ex- 
pression is  to  be  taken  as  a  factor. 

A  product  consisting  entirely  of  equal  factors  is  called  a 
power  of  the  repeated  factor.  The  repeated  factor  is  called  the 
base  of  the  power. 

E.g.  7^  means  x  •  x  •  x  and  is  read  the  third  power  of  x  or  x  cube ;  x'^ 
means  x  ■  x-  x  ■  x-  x,  and  is  read  the  fifth  power  of  x  or  briefly  x  fifth. 
{x  —  yY  =  (^  —  2/)(^  —  y)(^  —  y)  ^"^  '^^  r^^d  X  —y  cubed  or  the  cube 
of  the  binomial  x  —  y. 

The  first  power  of  x  is  written  without  an  exponent.  Thus  x 
means  x^ ;  2  means  2^  etc. 

134 


PRODUCT  OF  POWERS  OF  THE  SAME  BASE     135 

130.  Notice  the  difference  between  a  coefficient  and  an  ex- 
ponent. A  coefficient  is  a  factor^  while  an  exponent  shows 
how  many  times  some  number  is  used  as  a  factor. 

E.g.   5a  =  a  +  «  +  a  +  a  +  a,  while  a^  =  a-  a  -a-  a-  a. 

EXERCISES 
Perform  the  following  indicated  multiplications  : 

1.  23,  2*,  2^  5.    &,&.  9.    {a-\-h  +  c)\ 

2.  32,  33,  3^  6.    (a  +  &)l  10.    {a-\-h-c)\ 

3.  42,  43,  4^  7.    {c-d)\  11.    (3-a)2. 

4.  52,5^5*.  8.   982=  (100 -2)2.  12.    (3-6-c)2. 

PRODUCT  OF  TWO  POWERS  OF  THE   SAME  BASE 

131.  In  the  case  of  factors  expressed  in  Arabic  figures  mul- 
tiplications like  the  following  may  be  carried  out  in  either  of 
two  ways. 

E.g.         32 -3*  =  9 -81  =  729, 
or  32  .  34  =  (3  ■  3)(3  .3.3-3)=  32+*  =  3«  =  729. 

But  with  literal  factors  the  second  process  only  is  possible. 

E.  g.  a^  ■  a^  =  {a  '  a)(a  '  a  •  a  •  a)  =  a^+*  =  a*. 

EXERCISES 

In  the  following  exercises  carry  out  each  indicated  multipli- 
cation in  two  ways  in  case  this  is  possible : 

1.  5.52.           4.    7-73.            7.    x''X\  10.   23.22.2*. 

2.  5^'5\          5.   a^.a\          8.    t^  •  t*.  11.   3.32.33. 

3.  62.62.          6.   x^-a^.           9.    t''t'-t\  12.    22.23-22.2. 

132.  Illustrative  Problem.  To  multiply  2*  by  2",  k  and  n 
being  any  two  positive  integers. 

Solution.     2*  means  2  •  2  •  2  •  2,  etc.,  to  k  factors, 
and  2"  means  2  •  2  •  2  -2,  etc.,  to  n  factors, 

Hence,     2*  •  2«  =  (2  •  2  •  2  •  • .  to  A:  factors)  (2  •  2  ...  to  n  factors) 

-  2  .  2  .  2  •  2  ...  to  it  -I-  n  factors  in  all. 
That  is,  2*  •  2"  =  2*+"- 


136      PRODUCTS,  QUOTIENTS,  AND  FACTORS 

The  preceding  examples  illustrate 

Principle  XVI 
133.   Rule.     The  product  of  two  powers  of  the  same  base 
is  found  hy  adding  the  exponents  of  the  factors  and 
mahing  this  sum  the  exponent  of  the  cowimoji  base. 

Exponents  are  added  in  multiplication  only  when  they  apply 
to  the  same  base. 
.  E.g.  2'  •  32  =  8  •  9  =  72  cannot  be  found  by  adding  the  exponents. 

EXERCISES 

Perform  the  following  indicated  multiplications  by  means 
of  Principle  XVI: 

1.  2^'2K  5.  3*.3».  9.  52+".  52-". 

2.  a^  '  a\  6.  a-*  •  a;".  10.  a*"  •  a". 

3.  3^  •  3*.  7.  4"  .  4^  11.  c^ .  c^-*. 

4.  x*''a^.  8.  32«.32^  12.  of' -  x"". 

Perform  the  following  multiplications  by  means  of  Princi- 
ples II  and  XVI. 

13.  2^(22  +  2*).  16.   a\a^b-ab^). 

14.  22(3.2^  +  5.23).  17.   aP'^(4.af +  S  a^). 

15.  4*  (3  .  43  ~  5  .  42).  18.   V  (5  r"'  -  3  r«). 

Multiply  the  following : 

19.  aW  —  6  V  by  m*.  23.  S  y^^ -\-  4l  y*  —  y^  hj  y. 

20.  4. 32-5-7. 2by2*.  24.  7  a;^- 5  a^- 2  a;  by  a^. 

21.  4a;2_3aj3  +  6a5^by  a;2.  25.  S  a^b'-\-2  a^b-4:a^b^  hj  a\ 

22.  5a;-.3a;2^2a;*by  a;^  26.  4 a2«' - a^  +  02^^  by  a^*. 


PRODUCTS  OF  MONOMIALS 


137 


PRODUCTS  OF  MONOMIALS 

134.  In  multiplying  together  monomials  like  Sa^bc  and 
2  abhd  it  is  convenient  to  arrange  the  factors  so  that  the 
same  letters  are  associated  together  and  likewise  the  numerical 
coefficients.     This  we  are  permitted  to  do  by  Principle  XIII. 

Thus,  3  a%c  x  2  ab^cd  =  (3  •  2)(a2 .  a)(b  .  b^)(c  •  c)rf  =  6  a^b^c^. 

135.  Notice  that  in  the  product  the  eayponent  of  each  letter  is 
the  sum  of  the  exponents  of  this  letter  in  the  factors  and  the 
numerical  coefficient  is  the  product  of  the  numerical  coefficients 
of  the  factors. 

E.g.  (2  a262)(5  a^b^c)  =  10  a^+%^+^c  =  10  a%^c. 

This  is  a  convenient  rule  for  finding  the  product  of  two  or 
more  monomials. 

Ti/r    1.-    1  EXERCISES 

Multiply : 

1.  3a5by5a?62 

2.  4:a^hySxy\ 

3.  2  xyz  by  3  oc^yz. 

4.  6  A;'  by  7  ak^. 

5.  3  icy  by  4  xy^. 

6.  6  a-b^c  by  ab*c. 

7.  2  afb'c^  by  5  xb*c. 

8.  Sa^b*c  by  aM\ 

9.  2x(4:  +  7x*-3x'y). 

10.  4:yx(Sy^x'^-fx^-\-y*x*). 

11.  5  a262(a3_  53  ^^2^2) 

12.  4  icy (3  aa;  —  4  %  4-  2  a:?/). 

13.  (a  +  by(a-b). 

14.  (a-6)2(a+6). 
16.  (a-{-b)Xa-by. 


17.  (a4-&-c)(a  +  6  +  c). 

18.  (3aj-22/-l)(2a;-f2/). 

19.  (l+a  +  a2)(l-a). 

20.  (l-a  +  a2_^8)(i_^„)^ 

21.  (a-f  6)(a-6)(a2  +  62). 

22.  (a  +  6)(a2-a6  +  62). 

23.  (a-6)(a2_|-a6  4-6^. 

24.  (xi  +  x-\-l)(x^-x  +  l). 

25.  (x-\-y)(7^-x'y  +  xy'-f). 

26.  (a;-2/)(.x'«  +  a^2/  +  a:/+2/^. 

27.  (ar'-anz  +  ^/^Xa^-ha^  +  yO. 

28.  (a!2  +  2a;z/  +  y')(a;-2Z)^ 

29.  (x'-y^)(x*  +  xY  +  y'). 

30.  (ar'  +  2/2)(a;4_a;2^  +  2^). 

31.  (a^+y^(x-{-y)(x--y). 
16.    (a3  +  a26  +  a62  +  63)(^_^)     32.  (a;-y)(aj34-aJ2/+2/')(i«j'+y'). 


138  PRODUCTS,    QUOTIENTS,    AND   FACTORS 

QUOTIENT  OF  TWO  POWERS  OF  THE  SAME  BASE 

136.    Illustrative  Problem.     To  divide  x^  by  o^. 

Since  by  §  65  the  quotient  times  the  divisor  equals  the  dividend, 
we  seek  an  expression  which  multiplied  by  x'^  equals  x^. 

Since  by  Principle  XYI  two  powers  of  the  same  base  are  multiplied 
by  adding  their  exponents,  the  expression  sought  must  be  that  power 
of  X  whose  exponent  added  to  4  equals  6.  Hence  the  exponent  of  the 
quotient  is  6  —  4  =  2.     That  is,  x«  -f-  a;*  —  x^-^  =  x^. 

EXERCISES 

Perform  the  following  indicated  divisions: 


1. 

2^-22. 

8. 

5i-^-5'2. 

15. 

x'-^ar'. 

2. 

2' ^2'. 

9. 

724^722^ 

16. 

t''-^t\ 

3. 

2^-2. 

10. 

8^^-:- 8. 

17. 

m^  -7-  m. 

4. 

3^-f-3l 

11. 

6'-^6\ 

18. 

n^  -^  n\ 

5. 

3^  ^3. 

12. 

o?^o?. 

19. 

(20y  -  (20). 

6. 

3^  ^32. 

13. 

a'---a\ 

20. 

(101)14 --(101)". 

7. 

911-92. 

14. 

m'*  -T-  m^. 

21. 

(41)^-^(41)«. 

The  preceding  exercises  illustrate 

Principle  XVII 
137.   Rule.     The  quotient   of  two  jDoivers  of  the  same 
base  is   a  power   of   that    base  whose   exponent   is  the 
exponent  of  the  dividend  minus  that  of  the  divisor. 

For  the  present  only  those  cases  are  considered  in  which  the  expo- 
nent of  the  dividend  is  greater  than  or  equal  to  that  of  the  divisor. 
Notice  that  Principle  XVII  does  not  apply  to  powers  of  different 


E.g.  3'  -f-  2*  does  not  equal  any  integral  base  to  the  power,  7  —  4. 
This  division  can  be  performed  only  by  first  multiplying  out  both 
dividend  and  divisor. 


QUOTIENT  OF  POWERS   OF   THE   SAME   BASE  139 

EXERCISES 

Perforin  the  following  indicated  divisions  by  means  of  Prin- 
ciple XVII : 

1.  2^^2^  6.  ic"*" -T- a;^".  11.  ar''»+'' -^  a:«+\ 

2.  a'-r-o?.  7.  3-«-i--3«-2.  12.  w^^vf. 

3.  3^  ^31  8.  5»+^H-5''+2.  13.  (17)" --(17)13 

4.  y^^T?.  9.  x»+^-^a^H  14.  5a^-^2aj2. 

5.  33«-f-32».  10.  «*«-^<-.  15.  (12)* ---(12)3 

In  the  following  use  Principles  III,  V,  and  XVII : 

16.  (3.2*  +  5.23)-^2^  21.  {A:q^ -hx" -^-x'^)-^^. 

17.  (3.43_5.44)^42^  22.  (Sa^  4-9a*- 2a^)--a3. 

18.  (a^ft  -  a*62)  H- a^.  23.  iX^y^y-Worf^hx^)^^. 

19.  (4a:3^3a;4)^a^^.  24.  {7?^-^^-\-^-^^^-hx^.^'')^x^^^. 

20.  (a^m*  -  6  W) -V- m^  25.    {f^^^  +  y  •"-''- 'f)  ^  t- 

138.  The  process  of  division  by  subtracting  exponents  leads 
in  certain  cases  to  strange  results.  , 

Thus,  according  to  this  process,  3^-rrX*=x^^=x^,  which  is  as  yet 
without  meaning,  since  an  exponent  has  been  defined  only  when  it  is 
a  positive  integer.  The  exponent  zero  cannot  indicate,  as  in  the  case 
of  a  positive  integer,  how  many  times  the  base  is  used  as  a  factor. 
We  know,  however,  that  a;* -^x*  =  1,  since  any  number  divided  by  itself 
equals  unity.  Hence  if  we  use  the  symbol  x^  it  must  be  interpreted 
to  mean  1,  no  matter  what  number  x  represents. 

Again  by  this  process,  x^  ^  x^=x^~^=x~^,  which  is  as  yet  without 
meaning,  since  negative  exponents  have  not  been  defined.     But  we 

know  that  a;*  -^  a:*  =  ^  =  —  by  Principle  XV.     Hence  if  we  use  the 

symbol  x-^,  it  must  be  interpreted  to  mean  —     Negative  exponents 

x^ 
and  also  fractional  exponents  are  considered  in  detail  in  the  Ad- 
vanced Course. 


140      PRODUCTS,  QUOTIENTS,  AND  FACTORS 

DIVISION  OP  MONOMIALS 
139.   In  dealing  with  the   quotient   of  two  monomials  the 
indicated  division  may  be  written  in  the  form  of  a  fraction  and 
the  factors  common  to  dividend  and  divisor  may  be  cancelled 
by  Principle  XY  as  in  the  following  examples : 


(1) 

15xi«6V  -4- 

3  a%x''y  : 

3  a%x^y       x^y 

(2) 

12a^x-r-Zax: 

12  a% 
3  ax 

_4a 
1 

=  4a. 

Divide : 

EXERCISES 

1. 

4.7- 

9  by  2  .  3. 

5. 

Vl^y^\ij  ^xfz. 

2. 

12.8 

.  20  by  2  . 4 

.5. 

6. 

^a'h^^chjah'i^. 

3. 

Q^yh 

J  by  2  xyz. 

7. 

10x%^'c?hj2xWc. 

4. 

6*.  3^ 

.  ic^  by  62 .  52 

•  x\ 

8. 

36a;yby6iKy. 

9.  4tx'f-Zy?y''hyxY. 

10.  l^xY-l^^x^y^  +  Qx'y^hjQx'f. 

11.  49a*4-21a3-7aby  7a. 

12.  12  aa;y  —  16  aV?/^  +  8  a^xy  by  4  axy, 

13.  2a^4-4a;^'-8a^'*by  2af. 

14.  6  a^+i  4- 12  a^"+i  - 10  a;"+i  by  2  a;"+\ 

15.  4  a;^3  _  6  ^,115  _  10  a;*c  by  2  a;*. 

16.  10a»62-a*6«-hl5a*6^by  Sa^ftl 

FACTORS  OF  NUMBER  EXPRESSIONS 
140.  Factors  are  of  great  importance  in  arithmetic.  For  in- 
stance, the  multiplication  table  consists  of  pairs  of  factors 
whose  products  are  committed  to  memory  for  constant  use. 
Likewise  in  algebra  the  factors  of  certain  special  forms  of 
number  expressions  are  so  important  that  they  must  be  known 
at  sight. 


MONOMIAL  FACTORS  141 

Definition.  An  expression  containing  no  fractions  is  said  to 
be  prime  if  it  has  no  factors  in  the  integral  form  except  itself 
and  1. 

Thus  2,  3,  X,  X  +  2,  a^  -{-  b^,  are  prime  expressions. 

MONOMIAL  FACTORS 
141.  If  the  terms  of  a  polynomial  contain  a  common  monomial 
factor,  the  polynomial  may  be  divided  by  the  monomial,  and 
the  quotient  and  the  divisor  are  the  factors  of  the  polynomial. 
E.g.     a^  ~  ab  =  a  (a  —  b),    o  xy  —  S  x^y  -\-  4  x^y  =  xy(Q—  3  x  +  4:  a;^), 
and  6  3^y^^-{-  8  x^^hj^-  12  y^y^  =  2  ar'Y*(3  +  4  x'^+Y^  -  6  x^^y^). 

Observe  that  factoring  the  various  terms  of  a  polynomial 
does  not  factor  the  polynomial. 
■  E.g.    a^  -{■  ax  -\-  ab  +  by  is  not  factored  by  writing  it 
a(a-{-x)-^b(a  +  y). 
Likewise  10  a%'c  —  15  ab^c  +  20  abc^  is  not  factored,  although  each 
term   is  in  the  factored  form.      But  if  10  a%c  —  \o  ab^c  +  20  abc^  is 
written  in  the  form  5  abc  (2  a  —  3  6  +  4  c),  it  is  then  factored. 

Note  that  removing  a  common  factor  from  the  terms  of  a 
polynomial  is  nothing  more  than  the  application  of  Principle  I. 

EXERCISES 

Factor  the  following  polynomials : 

1.  7a2-f-14a6-h21a.  4.    9  ^w*  +  21  ?;*i(r»  -  18  vW. 

2.  13a*6-16a'6*-2a862.         5.   12  a*6' -  8  a'6*  -  6  a^ft*. 

3.  15V-20a:3y4-»2/.  6.   11  aV  -  44  a V  +  33  aa;. 

7.  72a;%''a-36x^6V-48aj363a^ 

8.  84«"6y  +  18a:36y-hl2a^6y. 

9.  VIa'bh^  +  ^la^Wc^-^^a%h\ 

10.  38  a^^jiv  _  76  a^ft^^^  -  76  a^^fe V. 

11.  4a;2Y'  +  6a^2/^'-8ar^«2/^. 

12.  3  a<'«+263«+<  +  6  a«''+*63"+3  _  12  a*«+3^5n+6^ 


142  PRODUCTS,    QUOTIENTS,   AND   FACTORS 

TRINOMIAL  SQUARES 

142.  In  §§89  and  90  we  found  by  multiplication: 

(a  +  6)2  =  a2  +  2a6  +  6^  (1) 

and  (a-by  =  a'-2ab-\-  b\  (2) 

By  means  of  these  formulas  we  may  square  the  sum  or  dif- 
ference of  any  two  number  expressions. 

E.g,    (3a:+7)2=(3x)2.f2(3a:).7  +  72^9a;2  +  42a:  +  49. 

(a  +  &+c)2=  [(«  +  &) +c]2=(a  +  ft)2  +  2(a  +  &)c  +  c2. 

[(5  +  r)-(s-0]2=(5  +  r)2_2(5  +  r)(5-0  +  (s-0'- 

The  last  expressions  may  now  be  reduced  by  performing  the 
indicated  operations. 

In  this  manner  write  the  squares  of  the  following  binomials. 
Read  as  many  as  possible  at  sight. 

1.  i  +  8.         3.    ^x-5y.  5.   m^  +  3n.         7.    Zf-^z^. 

:     2.  r-12.       4.    6a-7.  6.   l^  +  3y.       8.    %m''-ln\ 

9.  5a3-36^  12.    (3a-25)  +  5. 

10.  (a-3)-2(6+c).  13.    7a;-(4?'-s). 

11.  («-?/)  +  2(0  + 1/) .  14.    (m-  - 3)- (m^  +  n). 

143.  The  binomial  a  +  6  is  one  of  the  two  equal  factors  of 
a^  +  2  a6  +  h^,  and  is  called  the  square  root  of  this  trinomial. 

Likewise  a  —  6  is  the  square  root  of  a^  —  2  a6  +  &^- 

In  each  case  a  is  the  square  root  of  a?  and  h  of  Ir.     Hence 

2  ah  is  twice  the  product  of  the  square  roots  of  o?  and  b^. 

From  the  squares  obtained  in  the  last  article,  we  learn  to 

distinguish  whether  any  given  trinomial  is  a  perfect  square, 

as  in  the  following  examples : 

1.  a;2  +  4  a;  +  4  is  in  the.  form  of  (1),  since  z^  and  4  are  squares 
each  with  the  sign  +,  and  4  a;  is  twice  the  product  of  the  square  roots 
of  x^  and  4.     Hence 

x2  4-4a:  +  4  =  a:2+  2(2  a:)  +  22  =  (x -\-  2)(x  +  2)  =  (a;  +  2)2. 


TRINOMIAL   SQUARES  143 

2.  a:^  —  4  X  +  4  is  in  the  form  of  (2),  since  it  diifers  from  (1)  only 
in  the  sign  of  the  middle  term.     Thus 

a;2_4a;  +  4  =  a:2-2(2x)+22=  (x  -  2)(x  -  2)  =  (x -2y. 

3.  16  x^-^9  y^+l2  xy  is  not  in  the  form  (1)  since  12  xy  is  not  equal 
io2{^x){Zy). 

Definition.  A  trinomial  which  is  the  product  of  two  equal 
factors  is  called  a  trinomial  square. 

EXERCISES 

1.  State  the  facts  concerning  a  trinomial  which  make  it  a 
perfect  square. 

Determine  whether  the  following  are  trinomial  squares,  and 
if  so  indicate  the  two  equal  factors. 

2.  a?-\-2xy  +  y'^.  6.  m^  +  n^  — 2  w?i.  10.  a? -^r  h^  -  2  a%\ 

3.  x^-2xy-\-y^.  7.  7^-\-s^  +  2rs.  11.  ^-\-e-lQt. 

4.  a;^-f2ajy  +  ?A  8.  4ar^-8a^+4/.  12.  IQ  +  x^-Sx. 

5.  «*  -  2  a^y  +  ?/*.  9.  tt«  +  Z^"  +  2  a^'ft^  13.  9-6?/ +2/^. 

14.  25ar^  +  16?/4-40a;2/.  18.    16  a^  +  25  6^  _  50  a,6. 

15.  4m2  +  7i2  +  2?mi.  19.   16  a^ -f  25  6^  +  40  a6. 

16.  1004-s2+20s.  20.    81-270  6  +  225  62. 

17.  64  4-49  +  112.  21.    121 +  4  a26^-22  aft^. 

144.  From  the  foregoing  examples  we  see  that  a  tnnomial  is  a 
perfect  square  if  it  contains  two  terms  which  are  squares  each 
with  the  sign  +,  while  the  third  term,  whose  sign  is  either  -\-  or  —, 
is  twice  the  product  of  the  square  roots  of  the  other  two  Then  the 
square  root  of  the  trinomial  is  the  sum  or  the  difference  of  these 
square  roots  according  as  the  sign  of  the  third  term  is  +  or  —. 

Since  on  multiplying  we  find  (a  —  fe)^  and  (h  —  d)^  give  the  same 
result,  we  may  write  the  factors  oi  a^  —  2ab+  b^  either  (a  —  b)(a  —  h) 
or  (b  —  a)(b  —  a).  That  is,  the  square  root  of  (a  —  b)^  is  either 
a  —  ft  or  —  (a  —  6) .  Likewise  the  square  root  of  a^  is  either  +  a  or 
—  a.     See  page  176. 


144      PRODUCTS,  QUOTIENTS,  AND  FACTORS 

EXERCISES 

Factor  the  following.     If  any  one  of  the  trinomials  is  found 
not  to  be  a  square,  show  where  it  is  lacking. 

1.  9 +  2.  3.  4  +  16.    '  16.  121 -^  4.  a^-Ux\ 

2.  a;^  +  4  2/2  +  4i»2/.  17.  16  x^ -\- 64.  y^  -  64:  xy. 

3.  9x^-{-lSxy-\-9y\  18.  81  a^ -216  a  +  144. 

4.  4:x'-\-4,xy-^y\  19.  4.  a^ -j-S  ab^  +  4:b^. 

5.  4:Xr-^Sxy-^4:y\  20.  9  6*4- 186V  +  9  cl 

6.  25a^-f  12a;y+42/l  21.  4:X^  +  4:y^-Sxy. 

7.  16x'  +  16xy  +  4.y\  22.  9  a" -16  ab +  4:b\ 

8.  9  7-2  4-36  rs  +  25s2.  23.  9  a;^  -  24  a^ft  4- 16  ft^. 

9.  16a;«  +  8a;32/  +  2/'.  24.  25  4- 49  or*  -  70  a;. 

10.  4a:8  4-12a;V  +  9a^  25.  -  30  ab' -\- 9  a^ -^  25  b\ 

11.  a^«4-6a^&4-9  62.  26.  16  a^  -  24  aZ>  +  9  6-. 

12.  (a4-l)'4-2(a  +  l)&4-^>'.  27.  36  a;^  -  84  a;  +  49. 

13.  (a;4-3)2+4(a;+3)2/4-4  2/'.  28.  25-90  +  81. 

14.  a;«+12a^  +  36.  29.  64 a;^ -  32 a;  +  9. 

15.  a^  +  18a2  +  12.  30.  (3  +  ay-\-b'-2b(3+d) 

THE  DIFFERENCE  OF  TWO  SQUARES 
145.   In  §  91  we  found  by  multiplication, 
(a-{-b)(a-b)=a'-b'. 

By  means  of  this  formula  the  product  of  the  sum  and  differ- 
ence of  any  two  number  expressions  may  be  found. 

E.g.  (x  +  y-z)(x  +  y  +  z)  =  [(x  +  y)-  z]\_{x  4  2/)4  2] 
=  {x  +  yy-z'. 

In  this   manner  form  the  following  products.     Verify  the 
first  ten  by  actual  multiplication.    Eead  at  sight  the  first  seven. 

1.  (4a  +  56)(4tt-56).  4.   {5-6f){p +  6f). 

2.  (24a;  +  12?/)(24a;-12  2/).       6.   {3x-2y)(3x  +  2y\ 

3.  {16a^W-3c){16a?W-^3c).       6.   {^-f){:^  +  y^). 


THE  DIFFERENCE  OF  TWO  SQUARES  146 

7.  (Sx*-5x')(Sx*-^5y'). 

8.  lx-{-(:y-z)2[x-(y-z)l 

9.  (x""  -\-  y''){xr  —  2/") . 

10.  [c  -  (a -6)][c +(«-&)]. 

11.  lx-{y  +  z)-]lx  +  {y  +  z)-]. 

12.  (a-\-h-\-c){a—h  —  c). 

13.  {a-\-h-\-c){a  —  h  —  c). 

14.  (a-6  +  c)(a-6-c). 
16.  {r-y-z)(r-y  +  z). 
16.  {a^h  +  c){a-^h  —  c), 

146.  From  the  preceding  examples  we  see  that  evei^  binomial 
which  is  the  difference  between  two  perfect  squares  is  composed  of 
two  binomial  factoids;  namely,  the  sum  and  the  difference  of  the 
square  roots  of  these  squares. 

E.g.  IG  x^  —  9  ^2  is  the  difference  between  two  squares  (4  x)^  and 
(Syy.     Hence  we  have 

16  x2  -  9  y2  =  (4  xy  -  (3  y)2  =  (4  a;  +  3  y)(4  x  -  3  y). 

Again,(a-3i)2-(2a  +  6)2=[(a-3  6)  +  (2a+6)][(a-36)-(2a  +  &)] 
=  (a_3j  +  2a  +  6)(rt-36-2a-6) 
=  (3a-26)(-a-46). 

EXERCISES 

Factor  each  of  the  following.  Read  as  many  as  possible  at 
sight. 

1.  a^-4.y\  8.  a'-l.  15.  4.-(x-2yy. 

2.  9a^-36  2/'.  9.  l-9a?*.  16.  16a-25ab\ 

3.  x*-b\  10.  4:-36a\  17.  49x-4:xy\ 

4.  4a;2_9  58  ^^  l-64a8.  18.  225^6ix'y'. 

5.  16  a* -9  5*.  12.  144x2^-1.  19.  576  a -144  a/. 

6.  64-61  13.  256a*6«-c2.  20.  5«-3«. 

7.  1-61  14.  l-(x-\-yy.  21.  a;^-81/. 


146  PRODUCTS,    QUOTIENTS,   AND  FACTORS 

147.    It  is  important  to  determine  whether  a  given  expres- 
sion can  be  written  as  the  difference  of  two  squares. 

E.g.     a2  +  6-2  +  2  a6  -  c2  =  (a  +  by  _  c^  =  (a  +  &  +  c)(a  +  6  -  c). 
Also,  c2  -  a2  +  2  a&  -  62  =  c2  -  (a2  -  2  a6  +  62)  ^  ^^  -  (a  -  6)2 

=  (c  — a +6)(c  +  a  — 6). 
EXERCISES 

In  the  following  determine  whether  each  is  the  difference  of 
two  squares,  and  if  so,  factor  it  accordingly. 

1.  x'-iif-zf.  5.   4a262_(a2  +  62_c2)2. 

2.  {x-yy-z\  6.   a2-(624-c2  +  26c). 

3.  a^  +  V'-^ah-A.  7.   (2  a- 5^- (3  a +  1)2. 

4.  x'-\-y''.-2xij-z\  8.    {Sx^-yf-{x-\-yf, 

9.  (3a-25)2-(8a  +  5?>)2. 

10.  (3  m -4)2- (2  m +  3)2. 

11.  (2r  +  s)2-(3r-s)2. 

12.  81-(a+?>  +  c)2. 

13.  £c^  +  aj2  +  l-4a2. 

14.  a2_(^  +  2  2/)2. 

15.  9i»2-(a-&)2. 

16.  25m2-(3r  +  2s)2. 

17.  4c2-(4a24.i2a6-9  62). 

18.  9a2+6a6  +  52_c2.      • 

19.  16a;2-a2  4.4ct6_462. 

20.  a2-2a&  +  62_c2. 

21.  c2-(a2  +  2a5+62). 

22.  c2-(a2_2a6  +  62). 

23.  (a +  6)2 -(a -6)2. 
■      24.  a2  +  4a6  +  62_a^. 

25.  o?  +  4.ah-\-4th^-x'. 

26.  9a2  +  1662_-32a6-a;\ 


DRILL   EXERCISES  ,  147 

DRILL  EXERCISES 

1.  16c-(41-7c)  +  (15-8c). 

2.  _  (5  a  -  3  c)  -  (2  c  -  8  a)  4- 3  a. 

3.  -(-12x-7y-lox)-(-9y  +  Sx-\-3y), 

4.  (19  x-{-4:y  -32  x-17  x)  -12  X-  (^9  y  -\-lS  X-  70  x). 

5.  17a-3-(7a-2)4.(6a-5). 

6.  7(m  4-  6)  +  10  m  =  42  -  8(2  m  +  2)  + 181. 

7.  20-3(a;-4)4-2ir  =  2ic4-17. 

o        a;      ,     4rc        ^  f  5aj— 3?/=4— 2  £c+7  ?/, 

8.  ^4 r^  =  5-  13. 


a;-2      a;4-3 


(5x-3y=4: 
\5y  +  x  =  7. 


10. 


^4-a     »4-6_  1 
,2x  +  y=-i.  \.3x-2y  +  5z  =  31. 

[3x—y  =  D.  ieo,>i  ^ 

'^  ^  15.    j  3  a;  4-4  V  — 2=1, 

f3a;-7y=-ll,  o  .  o 

12.    "^  I— 2iB  — 2/4-3z  =  5. 


2  a;  4- 2/ =  4. 

16.  Represent  by  a  graph  the  progress  of  a  train  going  40 
miles  per  hour.  Is  the  distance  an  increasing  or  decreasing 
variable  as  the  time  increases? 

17.  Represent  by  a  graph  the  equation:  2x  —  3y  =  5. 
Does  y  increase  or  decrease  as  x  increases? 

18.  Graph  a;  4-  2  y  =  6  and  state  whether  y  increases  or  de- 
creases as  x  increases. 

19.  How  do  you  represent  a  fraction  whose  numerator  is  3  less 

than  twice  that  of  —  and  whose  denorfiinator  is  equal  to  3  times 
n 

the  sum  obtained  by  adding  2  to  the  denominator  of  this  fraction  ? 

20.  There  are  two  numbers  such  that  if  one  half  their  prod- 
uct is  divided  by  twice  their  sum  the  result  is  12  times  their 
difference.  Write  an  equation  representing  this  relation  be- 
tween the  numbers. 


148  PRODUCTS,   QUOTIENTS,   AND  FACTORS 

DIVISION  BY  A  POLYNOMIAL 

148.  The  simplest  case  of  division  by  a  polynomial  is 
that  in  which  the  dividend  can  be  resolved  into  two  factors, 
one  being  the  given  polynomial  divisor  and  the  other  a 
monomial. 

E.g.  To  divide  4  a;^  -f  4  x^y  by  a:  +  y,  factor  the  dividend,  and  we 
have  4  x^(x  -\-y)-^(x  +  y)  =4  x^. 

In  case  the  dividend  cannot  be  factored  in  this  man- 
ner, then,  if  the  division  is  possible,  the  quotient  must  be 
a  polynomial.  The  process  of  finding  the  quotient  under 
such  circumstances  is  best  shown  by  studying  a  particular 
case. 

Illustrative  Example  1.     Consider  the  product 

(x2  -\-2xy+  y^)(x  +  y)  =  x^(x  +  y)  +  2  xy{x  +  y)  +  y\x  +  y). 

The  products,  x\x  +  ?/),  2  xy{x  +  y),  and  y'^{x  4-  y)  are  called 
partial  products,  and  their  sum,  a;^  +  3  x^y  +  3  xy^  +  y^^  the  complete 
product. 

In  dividing  a;^  +  3  x'^y  +  3  xy^  +  y^  by  x  +  y  the  quotient  must  be 
such  a  polynomial  that  when  its  terms  are  multiplied  hy  x  -{-  y  the 
results  are  these  partial  products,  which  in  the  solution  are  called 
1st,  2d,  and  3d  products. 

The  work  may  be  arranged  as  follows : 
Dividend  or  product :  a:^  +  3  x'^y  +  3  xy^  -^  y^    ^  +  y,  divisor. 

1st  product,  x^(^x  +  y)  :  x^  -\-     x^y  x^  -h  2xy  -\-  y% 

Dividend  minus  1st  product :  2  x^y  +  3  xy^  +  y^        [quotient. 

2d  product,  2xy(x  -{-  y):  2  x^y  +  2  xy^ 

Dividend  minus  1st  and  2d  products  :  xy^  +  y^ 

3d  product,  y^(x  +  y)  :  xy^  +  y^ 

Dividend  minus  1st,  2d,  and  3d  products :  0 

Explanation.  Since  the  dividend  or  product  contains  the  term  a:^, 
and  since  one  of  the  factors,  the  divisor,  contains  the  term  x,  the 
other  factor,  the  quotient,  must  contain  the  term  x^.     Multiplying 


DIVISION  BY  A  POLYNOMIAL  149 

this  terni  of  the  quotient  by  the  divisor,  we  obtain  the  first  partial 
product,  z^  +  x^y. 

Subtracting  the  first  partial  product  from  the  whole  product 
a;^  4-  3  x^y  +  3  xy^  +  y^,  the  remainder  is  2  x^y  +  3  xy^  +  y^,  which 
is  the  product  of  the  divisor  and  that  part  of  the  quotient  which 
has  not  yet  been  found.  Since  this  product  contains  the  term 
2x^y  and  the  divisor  contains  the  term  x,  the  quotient  must  contain 
the  term  2  xy.  The  product  of  2  xy  and  a:  -f  y  is  the  second  partial 
product. 

In  like  manner  the  third  partial  product  is  xy^  +  y*. 

Subtracting  the  third  partial  product  the  remainder  is  zero. 
Hence  the  sum  of  the  three  partial  products  thus  obtained  is  equal 
to  the  whole  product,  and  it  follows  that  x^  -\-  2  xy  -{-  y^  is  the 
required  quotient. 

149.  Problems  in  division  may  be  checked  by  substituting 
any  convenient  values  for  the  letters.  For  example,  in  this 
case,  x  =  l,  y=l,  reduces  the  dividend  to  8,  the  divisor  to  2, 
and  the  quotient  to  4,  which  verifies  the  correctness  of  the 
result. 

Since  division  by  zero  is  impossible  (see  Advanced  Course), 
care  must  be  taken  not  to  select  such  values  for  the  letters 
as  Avill  reduce  the  divisor  to  zero. 

Illustrative  Example  2.  Divide  2a;*4-ic'  —  7a^-f5ic  —  1  by 
x^-\-2x-l. 

Solution.  [divisor. 

Dividend  or  product :  ]2x*+    x^-7 x^-\-5x-l       a:'^+2x-l, 

1st  product,  2  x\x^  +  2  x  -  1) :     2x*  +  4x^-2x^ 2x^-dx  +  l, 

Dividend  minus  1st  product:  —Soc^  —  bx^  +  ox  —  l      [quotient. 

2d  product,  -  3 x(x^  +  2 x  -  1)  :        -^x^-6x^  +  3x 
Dividend  minus  1st  and  2d  products  :  +x*+2a;  — 1 

3d  product,  1  •  (a;2  +  2 ar  -  1)  :  ar''+2a:-l 

Dividend  minus  1st,  2d,  and  3d  products:  0 

Check.    Substitute  a;  =  2  in  dividend,  divisor,  and  quotient. 


150  PRODUCTS,    QUOTIENTS,    AND   FACTORS 

Illustrative  Example  3.     Divide  20  a^— 4  + 18  a*+ 18  a  — 19  a^ 
by  2tt2-3a-f-4. 

Solution.     Arranging  dividend  and  divisor  according  to  the  de- 


&r                     > 

[divisor. 

Dividend  or  product :             18  a^- 19  a^  +  20  aH  18  a  -  4 

2a2-3a+4, 

1st  product:            .                  liia'^-21  a^+ma^ 

9a2-|-4a-2, 

Dividend  minus  1st  product :             8  a^- 16  a2_|_  is  a  -4 

[quotient. 

2dj)roduct:                                             8a8-12a2+16a 

Dividend  minus  1st  and  2d  products  :       —^a^-\-  2  a  — 4 

3d  product:                                                    -4a2+  6a-8 

Dividend  minus  all  products :  —  4a  +  4 

Since  2  a^  is  not  contained  in  —  4  a,  the  division  ends  and  —  4  a  +  4 
is  the  remainder.  As  in  arithmetic  we  write  this  as  the  numerator 
of  a  fraction  whose  denominator  is  the  divisor.     Hence,  the  complete 

result  is  9  a2  +  4a  -  2  + i-^lo 

2a2-3a  +  4 

Check.     Substitute  a  =  1  in  dividend,  divisor,  and  quotient. 

150.  From  a  consideration  of  the  preceding  examples  the 
process  of  dividing  by  a  polynomial  is  described  as  follows : 

1.  Arrange  the  terms  of  dividend  and  divisor  according  to 
descending  (or  ascending)  powers  of  some  common  letter. 

2.  Divide  the  first  term  of  the  dividend  by  the  first  term  of 
the  divisor.     This  quotient  is  the  first  term  of  the  quotient. 

3.  Multiply  tbe  first  term  of  the  quotient  by  the  divisor 
and  subtract  the  product  from  the  dividend. 

4.  Divide  the  first  term  of  this  remainder  by  the  first  term 
of  the  divisor,  obtaining  the  second  term  of  the  quotient. 
Multiply  the  divisor  by  the  second  term  of  the  quotient  and 
subtract,  obtaining  a  second  remainder. 

5.  Continue  in  this  manner  until  the  last  remainder  is  zero, 
or  until  a  remainder  is  found  whose  first  term  does  not  contain 
as  a  factor  the  first  term  of  the  divisor.  In  case  no  remainder 
is  zero,  the  division  is  not  exact. 


DIVISION  BY   A   POLYNOMIAL  151 

EXERCISES 

Check  the  result  in  each  case,  being  careful  to  substitute 
such  numbers  for  the  letters  as  do  not  make  the  divisor  zero. 
Divide  the  following : 

1.  a^-^2ab-{-b'hy  a-h b, 

2.  a''-2ab-\-b'hya-b. 

3.  a^-S  a'b  +  3  ab^  -b^hy  a-b, 

4.  2 a;^  +  2 ai^y  —  4=0^  —  x— 4:xy  —  y  by  x-\-y. 

5.  a:^  -\-  xy^  —  a^y  —  y^  by  X  —  y. 

6.  x^  +  4:a^  +  x-6hy  x-^3. 

7.  a^  +  4a^  +  a;— 6bya;  —  1. 

8.  x^-6a^-\-2a?-Sx  +  6hyx-'l. 

9.  a^  +  3x'y-\-Sxf-\-fhyx'-^2xy-^f. 

10.  a^-Sx"  -\-75hyx-5. 

11.  2  a^  +  19  a'b  -[- 3ab^  by  2  a -\-b. 

12.  X*  —  ^x^y  -{-  6  x^y-  —  4  xy^  -f-  y*  by  x  —  y. 

13.  a;^  +  4  x^y  +  6  xY  +  4:xf  -\-y*  by  x^  -\- 2  xy  -\-  y\ 

14.  X*  +  ^y  -\-  xy^  -\-  y*  by  x  -\-  y. 

15.  x*-\-x^y^  +  t/by  x^-xy-^  y\ 

16.  X*  —  y*by  X  —  y. 

17.  a^  +  b^  +  (^  -  3  abcby  a  -^  b  -\-  c. 

18.  2  X* -^llx^- 26  a^  +  Ux  + 3  by  a^-\-7x- 3. 

19.  a5«+5a;^yH-10a:3^2_^10a^?/3^5a,y4_,_yj  ]^y  x^-\-2xy-\-f. 

20.  ar^  -  a;^  -  27  ic^  ^  10  0^2  _  30  3.  _  200  by  ic2  -  4  a;  -  10. 

21.  3  a^  -  4  a;?/  +  8  a^z  -  4  /  +  6  z/2  4-  3  ^2  by  a;  -  2  2/  +  3  2. 

22.  9  r^s^  -  4  r^^^  _^  4  ,,5^2  _  ^^2  \)y3rs-2rt-h  st. 

23.  9  d'b''  -{- 16  x"  -  ^a" -36  b^'x'  by  3  ab  -\-  6  bx  -  2 a  -  4.x. 

24.  »^  +  a^2/  -\-x^z  —  xyz  —  yh  —  yz^  by  ar^  —  yz, 

25.  (i^  +  a'^>  +  a^  -  a^ft^  _  2  ad^  +  6^  by  a^  +  a6  -  6^. 

26.  a^  +  6^  +  3  a6  - 1  by  a  +  6  - 1. 

27.  a^*  -  3  a2*6*  +  3  a*^^*  -  5^*  by  a*  -  6*. 


152   .   PRODUCTS,  QUOTIENTS,  AND  FACTORS 

THE  SUM  OP  TWO  CUBES 
151.    Example.     Divide  a^  -{- b^  by  a -{■  b. 
Since  the  quotient  multiplied  by  the  divisor  equals  the  divi- 
dend we  have,     ^3  _^  ^3  ^  („  ^  ^)(^2  _  ^^  ^  ^2^^ 

This  shows  that  every  binomial  which  is  the  sum  of  two  cubes 
is  the  product  of  two  factors,  one  of  which  is  the  sum  of  the 
numbers,  and  the  other  is  the  sum  of  their  squares  minus  their 
product. 

E.g.  (1)  x^-]-y»=(x-h  y)  (x2  -  xy -^  y^) . 

(2)  8a*  +  27  63  =  (2a)3  +  (3&)8 

=  (2  a  +  3  &)(4  a2  -  2  a  .  3  5  +  9  &2). 

(3)  x6  +  2/6  =  (x^y  +  (2,2)8  =  (a:2  +  y^)  (x\ -  xY  +  y*). 

Notice  the  difference  between  the  trinomial  x^  —  xy  -{-  y^  and  the 
trinomial  square  x^  —  2xy  +  y^. 

EXERCISES 

Determine  whether  each  of  the  following  is  the  sura  of  two 

cubes,  and  if  so,  find  the  factors.     Check  the  results  by  multi- 
plication. 

1.  a^-^f.  10.    Sa^  +  27b\  19.    64.x^-{-27f. 

2.  a»  +  8  6^  11.   8a3  +  64&3.  20.   S^  +  IO^ 

3.  27  a^  +  bK  12.   w^x^  +  a^aK  21.   l  +  729x\ 

4.  Sa^  +  l.  13.   1+S  a^b\  22.    a^-\-f\ 

5.  1  + 64  0^3^  14.    64a^  +  343.  23.    a^-{-b\ 

6.  2=^  +  33,  15.    l-\-a\  24.   27r3  +  125s3. 

7.  125  +  729.  16.    a^-^9b\  25.   x^-^27y\ 

8.  14-I25ic«.  17.    125  0^  +  2^.  26.   64  +  a». 

9.  27a;«  +  l.  18.    l+a».  27.    a«6«  +  a^/. 

•  28.  Find  whether  a:^  4- ?/^  is  exactly  divisible  by  x  —  y',hj 
x^  —  y^;  by  x^-\-y'^.  What  binomial  is  a  divisor  of  the  sum 
of  two  cubes  ?     What  is  the  quotient  ? 


THE  DIFFERENCE   OF  TWO   CUBES  153 

THE  DIFFERENCE  OF  TWO  CUBES 

152.   Example.     Divide  a^  —  b^hy  a  —  b  and  obtain 
a^-b^  =  (a-  b)  {a'  +  ab  +  b') 

This  shows  that  the  difference  of  the  cubes  of  two  numbers  is 
the  product  of  two  factors^  one  of  which  is  the  difference  of  the 
numbers,  and  the  other  the  sum  of  the  squares  of  the  numbers  plus 
their  product. 

E.g.  (1)  x^-y^=(x-  y)(x2  +  xy  +  rf). 

(2)  8a8-646»=(2a)3-  (4&)8 

=  (2a-46)(4a2  +  2a.46+  1662). 

(3)  a9  -  69  =  (a8)8-  (63)3^  (^8  -  ¥){a^  +  a^ft'  +  b^). 

Notice  the  difference  between  the  factor  x^  -\-xy-\-  y^j  and  the  tri- 
nomial square  x^  -\-2xp-\^  y^. 

EXERCISES 

Determine  whether  each  of  the  following  is  the  difference  of 
two  cubes,  and  if  so,  find  the  factors.  Check  the  results  by 
multiplication. 

1.  7^-^.  9.  27- 125  a^ 

2.  Sa^-b\  10.  a^-if. 

3.  a^-Sb\  11.  l-x\ 

4.  Sa^-Sb^  12.  a^-S. 

5.  33-28.  13.  l-125a^. 

6.  i-a^^  14.  8-27a^. 

7.  l-8a3.  15.   27a^-64. 

8.  64a^-^.         16.   2V-1. 

25.  Also  factor'  Exs.  10, 11,  and  20  as  the  difference  of  two 
squares. 

26.  Find  whether  a^  —  y^  is  exactly  divisible  by  x-^y,  by 
a^  +  /,  by  a;2  —  y^.  What  binomial  is  a  divisor  of  the  difference 
of  two  cubes  ?    What  is  the  quotient  ? 


17. 

Sx*-f. 

18. 

64  a^  -  27  b\ 

19. 

l-729aJ». 

20. 

0^-2/^^ 

21. 

27  7^- 125  s3. 

22. 

125a^-Sly\ 

23. 

729-16aj3. 

24. 

125a«-64  6^ 

154  PRODUCTS,    QUOTIENTS,   AND  FACTORS 

TRINOMIALS  OF  THE  FORM  X^ -^  (a  +  b) X -{- ab 

153.  In  §  92  were  found  such  products  as 

(1)  (x-\-5){x-\-2)  =  x'-{-7x-\-10, 

(2)  (x-5J(x-2)  =  x'-7x-\-10, 

(3)  (x  +  5Xx-2)=a^  +  Sx-10. 

(4)  (x-5)(x  +  2)  =  3^-Sx-10, 

All  these  are  included  in  the  form 

(x  +  a){x  +  b)  =  x'  -\-(a  -h  b) X  -\-  aby 

in  .which  the  coefficient  of  x  is  the  algebraic  sum  of  a  and  b  and 
the  last  term  is  their  product. 

154.  It  is  possible  to  recognize  such  products  at  sight,  and 
thus  to  find  the  factors  by  inspection. 

Illustrative  Examples.  Determine  whether  the  following  tri- 
nomials are  of  the  kind  just  considered : 

1.  ar^  -f  7  aj  -h  12.  The  question  is  whether  two  numbers  can 
be  found  such  that  their  sum  is  +  7  and  their  product  12.  The 
numbers  3  and  4  answer  these  conditions.     Hence, 

x'  +  7x  +  12  =  (x  +  3){x  +  4.). 

2.  o;^  — 5  a;— 14.  Since  the  product  of  the  numbers  sought 
is  — 14,  one  number  must  have  the  sign  —  and  the  other  +  ; 
and  since  their  sum  is  —  5,  the  one  having  the  greater  absolute 
value  must  have  the  sign  — .  Hence,  the  numbers  are  —  7  and 
+  2,  and  we  have  a^  -  5  a;  —  14  =  (a;  —  7)(a;  +  2). 

3.  a;2_7a;  +  i2  =  (a;-3)(a;-4).     Since   (-3)(-4)=-f-12 

and(-3)  +  (-4)  =  -7. 

4.  x'  +  4.x-12=(x  +  6)(x-2).  Since  (-|-6)(- 2)=  -  12 
and(+6)  +  (-2)  =  +  4. 

It  is  to  be  noted  that  it  is  not  always  possible  to  find  inte- 
gers to  fulfill  these  two  conditions. 


TRINOMIALS  OF  THE  FORM  a^ +  (a -f-6)a;  + a6    '   155 

E.g.    Given  x^  +  5  a;  +  3.     By  inspection,  it  is  easily  seen  that  there 
are  no  two  integers  such  that  their  sura  is  +  5  and  their  product  +  3. 

Tell  the  steps  to  be  taken  in  deciding  whether  x^-\-ax-\-h 
has  two  binomial  factors. 

EXERCISES 

Determine  whether  each  of  the  following  trinomials  can  be 
factored  by  inspection,  and  if  so,  find  the  factors. 

26.  a^- 14  a -51. 

27.  a2-3a-54. 

28.  a;^-8  0^^-32. 

29.  a«-3a3-154. 

30.  x2-10a;-f25. 

31.  a^¥-l^ay-m. 

32.  ^-llxyz  +  12fz\ 

33.  9-«  +  6r*s-91s2. 

34.  aV  +  9aV-162. 

35.  a^-f  11  a -210. 

36.  m*  +  4  m V  H- 4  71^ 

37.  ^t^-l^st-5^. 

38.  a^ft^  _  27  a6H- 26. 

39.  Z^ 4. 13  24.42. 

40.  a?f.-llxy-1^0, 

41.  9a2  +  24a  +  16. 

42.  81  a^- 99  a +  30. 

43.  ^2.^26  9^  +  133. 

44.  ar^  +  5jci/— 84  2/1 

45.  r2  +  3r-154. 

46.  u^-\-Z^uv  +  lQbv^. 

47.  (a +  6)2 -19  (a +  6) +  88. 

48.  (a;_2/)2-14(a;-2/)  +  40. 
.49.  (r-s)2-17(r-s)  +  60. 

50.   x^  +  {a-{-h)x-\-ah. 


1. 

ar^  +  3aj  +  2. 

2. 

x^  +  x-Q. 

3. 

x'-x-Q. 

4. 

ic2_6a;  +  8. 

5. 

a:2  +  6a;  +  8. 

6. 

a^_3a;_8. 

7. 

ar^  4.2^-8. 

8. 

a2_4a-32. 

9. 

a2  +  4a-32. 

10. 

52.^156  +  56. 

11. 

6=^  +  86  +  15. 

12. 

62_5_56. 

13. 

l^^jj-m. 

14. 

c2_3c-15. 

15. 

x'-l^x  +  m. 

16. 

ic2  +  15a;-54. 

17. 

ic2_14a;-95. 

18. 

2/2 +  21 2/ +  98.^ 

19. 

2/'-72/-98. 

20. 

a^_19a;  +  78. 

21. 

a;*  +  18a;2  +  77. 

22. 

a;*_5a2-104. 

23. 

a2  +  32a  +  240. 

24. 

a*_lla2  +  28. 

25. 

a*_lla2-60. 

156  PRODUCTS,   QUOTIENTS,   AND  FACTORS 

TRINOMIALS   OF   THE  FORM  aX^-{-bx  +  G 

155.  Find  the  product  of  2  a;  -j-  5  and  3  x-\-2. 

2x  +  5 
Sx-^2 
6a;2  +  15aj 

4a;-fl0 

60^  +  19  a;  +  10 

The -products  3  a;  •  2  a;  and  2  •  5  are  called  end  products  and 
2  •  2  a;  and  5  •  3  a;  are  called  cross  products.  In  this  case  the 
cross  products  are  similar  with  respect  to  x  and  are  added. 
Hence  the  Jinal  product  is  a  trinomial  two  of  whose  terms  are  the 
end  products  and  the  third  term  is  the  sum  of  the  cross  products. 

This  fact  enables  us  to  write  such  products  at  once. 

E.g.    (3  a  +  4)(5  a  -  7)  =  15  a2  -  a  -  28. 

In  this  case  15  a^  is  the  first  end  product  and  —  28  the  second,  while 
—  a  is  the  sum  of  the  two  cross  products,  20  a  and  — 21  a. 

EXERCISES 

In  this  manner  obtain  the  following  products : 

1.  (2a  +  3)(a  +  3).  13.  {Sx-2y)(x  +  Sy). 

2.  (4a-l)(3a  +  2).  14.  (4:a-3y)(a  +  y). 

3.  (2a;  +  5)(a;-7).  15.  (3r~2s)(2r  +  s). 

4.  (7r  +  8)(3r-6).  16.  (5m -n)(2m  +  7t). 

5.  (2a;  +  8)(9a;-4).  17.  (5a  +  3a;)(3a-4a;). 

6.  (3m-l)(4m+3).  18.  (4a-5  6)(a  +  3  6). 

7.  (5s-7)(2s-4).  19.  (3  a -\- 5  b)(a  -  b). 

8.  (2a;-l)(7a;  +  4).  520.  (3  c- 7  d)(2c  +  3(«). 

9.  (4?i-9)(5n-7).  21.  (2a-36)(3a  +  2&). 

10.  (Sy-l){5y  +  ll).  ■    22.    {^x-5y){2x^-3y). 

11.  (f_5)(«+4).  23.    (5a  +  36)(2a-56). 

12.  (5  a;  — 2/) (2  a;  — 32/).  24.    (7m-f  5n)(3m  +  4w). 


TRINOMIALS   OF   THE   FORM   a3cr-{-bx-{-c  157 

156.  Trinomials  in  the  form  of  the  above  products  may  some- 
times be  factored  by  inspection. 

Ex.  1.     "Factor:  5a^  +  16a;4-3. 

If  this  is  the  product  of  two  binomials  they  must  be  such  that  the 
end  products  are  5  x^  and  3  and  the  sum  of  the  cross  products  16  x. 

One  pair  of  binomials  having  the  required  end  products  is  5  a:  +  3 
and  X  -\-  1,  others  are  5  a:  —  1  and  a:  -  3,  5  a;  +  1  and  a;  +  3,  and  ox  —  ^ 
and  a;  —  1. 

It  is  convenient  to  write  down  these  possible  paii's  of  factors  as 
follows,  as  if  arranged  for  multiplication : 

5a:  +  3  5a;-l  5a:  +  l  5a:-3 

a;  +  l  a:-3  a;  +  3  a;-l 

The  sum  of  the  cross  products  in  the  first  pair  is  8  x,  in  the  second 
pair  —  16  X,  in  the  third  pair  16  a:,  and  in  the  fourth  —  8  a;.  Since 
16  x  is  the  middle  term  required,  the  factor's  are  5  a;  +  1  and  a;  +  3. 

Ex.  2.    Factor:  6a2-5a-4. 

In  this  case,  as  in  the  one  preceding,  there  are  several  pairs  of  bi- 
nomials whose  end  products  are  6  a^  and  —  4,  such  as  2  a  —  2  and  3  a  +  2, 
6  a  —  1  and  a  +  4,  etc.  By  trial  we  find  that  among  these  3  a  —  4  and 
2  a  4-  1  is  the  only  pair  the  sum  of  whose  cross  products  is  —  5  a. 
Hence  6  a^  -  5  a  -  4  =  (3  a  -  4)  (2  a  +  1). 

In  this  manner  factor  the  following : 

1.  Sx^+5x-{-2.  11.  5a;- +  26 a; -24.  21.  12  c^-f  25  c+12. 

2.  9a2  +  9a  +  2.  12.  2a^-5a;  +  2.  22.  8-|-6a-5a2. 

3.  2a;2  +  llaj  +  12.  13.  2m^-m-S.  23.  15-5x-10a:2 

4.  9ar»  +  36a;-f-32.  14.  Tc'-Sc-A.  24.  6-5x-4.a^. 

5.  2a;2_^_28.  15.  5a;^  +  9a^-18.  25.  Sh' -ISh-^U. 

6.  12s2.+  lls4-2.  16.  7a^+123a2-54.  26.  15r'-r-2. 

7.  6e-  +  7t-3.  17.  6c2-19c  +  15.  27.  2«=^  +  ll^  +  5. 

8.  6a^-x-2.  18.  3a2-21a  +  30.  28.  10 - 5 a; - 15 ic^, 

9.  5r2  +  18r-8.  19.  6(^  +  4d-2.  29.  5a^ -SSx-{-lS. 
10.  14a2-39a+10.  20.  20  a^  -  a  -  99.  30.  20  -  9  a;  -  20  ar^. 


158  PRODUCTS,    QUOTIENTS,   AND   FACTORS 

FACTORS  FOUND  BY  GROUPING 

157.  Another  method  of  general  application  is  here  applied 
to  polynomials  of  four  terms. 

Ex.  1.     Find  the  factors  of  ax  -\-  ay  -\-hx-\-  by. 

By  Principle  I,  the  first  two  terms  may  be  added  and  also 
the  last  two. 

Thus,         ax  -\-  ay  -\-  hx  -\-  by  =  a{x  -\-  y)  -\-  h  {x  -^  y). 

These  two  compound  terms  have  a  common  factor,  (x-\-y), 
and  may  be  added  with  respect  to  this  factor  by  Principle  I. 

Thus,  a{x  +  y)+b(x-{-y)=(a  -f  b){x  +  y). 

Hence,  ax-\-ay-\-bx-\-by  =  {a  +  b)(x-\-y). 

Ex.  2.     Factor  ax  —  ay—  bx  +  by. 

Combining  the  first  two  terms  with  respect  to  a  and  the 
second  two  with  respect  to  —  b,  we  have, 

ax  —  ay  —  bx  -\-  by  =  a(x  —  y)  —  b{x  —  y). 

Again  combining  with  respect  to  the  factor  x  —  y, 
ax— ay  —  bx  -\-by  =  {a  —  b)(x  —  y). 

The  success  of  this  method  depends  upon  the  possibility  of 
so  grouping  and  combining  the  terms  as  to  reveal  a  common 
binomial  factor. 

EXERCISES 

Factor  the  following : 

1.  ab'^-{-ac'-db^-dc\  11.  2n^ -cn  +  2nd-cd. 

2.  6ms-15w^  +  9ws  — lOm^.  12.  bax  —  l^ay  —  ^bx-\-^by. 

3.  ^  ax  —  10  ay  -{-  4.bx  —  ^  by.  13.  3  xa  —  12  a;c  —  a  +  4  c. 

4.  2  a^-}-3  ak  — 14  an  —  21  nk.  14.  3xy  —  A  mn  -j-  6  my  —  2  xn. 

5.  ac-\-bc-\~  ad  +  bd.  15.  7  mn  -\-7  mr  —  2n^  —  2  nr. 

6.  ax^ —hx^—ay^-\-by^.  16.  a  —  l-{-a^  —  a^. 

7.  8ac-20a(^-66c  +  156d.  17.  3s+ 2  +  6s*-h4  s^. 

8.  2ax  —  ^bx-{-^by  —  ay.  18.  as^  —  3bst  —  ast  +  3  bt\ 

9.  5  4-  4  o  — 15  c  —  12  ac.  19.  3  mri  +  6  m^  —  2  am  —  an. 
10.    156-6-206c  +  8c.  20.  2  ar -^2  as  +  2br  +  2bs. 


SQUARES   OF  POLYNOMIALS  159 

SQUARES  OF  POLYNOMIALS 

158.  Example.  By  multiplication  find  the  square  of  a  +  6  +  c, 
and  reduce  the  result  to  simplest  form. 

How  many  terms  are  there  in  the  product  ?  How  many  are 
squares  ?     How  many  are  of  the  type  2  ab? 

From  this  we  get  the  following  rule : 

The  square  of  a  trinomial  consists  of  the  sum  of  the  squares  of  its 
terms  plies  twice  the  product  of  each  term  by  each  succeeding  term. 

In  symbols  this  is 

{a  +  b  +  cy=  a^  +  b'  4-c"-|-  2  ab +  2ac-{-2  be. 

EXERCISES 

1.  How  may  the  above  rule  be  applied  to  find  the  square 
of  a— 6-1- c,  ofa-t-6  —  c,  of  a  — b-\-c? 

Suggestion,     a  —  b-\-c  =  a-{-(—  b)+c. 

2.  Find  the  square  of  (2  a  -h  6  -  3  c). 

Suggestion.  (2  a  +  b  -  3  c)^  =  (2  a)^  +  b^  +  (- 3  c)2  +  2(2  a)l  f  2 
(2  a)(-  3  c)  +  2  .  &(  -  3  c).     This  should  now  be  simplified. 

Find  the  squares  of  the  following : 

3.  a-h26-f-3c.         6.   2x-\-Sy-z.  9.    5a-\-2b-Sc. 

4.  2a-h6  +  4c.         7.   4cX  —  y-{-2.  10.    a  —  b  —  2c. 

5.  x  —  Sy  —  z.  8.   3  a  — 6-f-c.  11.   x  —  Sy  —  z. 

12.  By  multiplication  find  the  square  of  a  +  bA-c-\-dj  and 
reduce  to  simplest  form.  Study  this  product  and  try  to  make 
a  rule  for  squaring  any  polynomial. 

159.  In  the  preceding  exercises,  the  square  of  each  trinomial 
consists  of  six  terms,  namely  three  squared  terms  and  three  sums 
of  cross  products. 

It  follows  that  the  square  root  of  such  a  polynomial  may  be 
found  by  taking  the  square  roots  of  the  three  squared  terms 
and  determining  their  signs  by  trial  in  such  a  way  as  to  give 
the  proper  cross  products. 


160  PRODUCTS,   QUOTIENTS,   AND   FACTORS 

Ex.  Find  the  square  root  of 

4:a^-12xy-16xz  +  9/  +  242/^4- 16  z\ 

Solution.  The  terms  4  x^,  9  y^,  and  16  s^  are  all  squares.  The  square 
root  of  4  x^  is  either  +  2  a:  or  —  2  x,  that  of  9  y^  is  either  -{-  3  y  or 
—  3  2/,  that  of  16  22  is  +  4  2  or  —4  2;.  Hence  if  the  given  polynomial 
is  a  perfect  square,  the  terms  —  12  xy,  —  16  xz,  and  24  yz  must  be  the 
sums  of  the  cross  products  of  2  x,  3  y,  and  4  z  each  taken  with  the 
proper  sign.  By  inspection  ,we  soon  find  that  the  sign  of  2  a;  must  be 
+ ,  and  that  oi  3  y  and  4  z  each  —  . 

Hence  the  required  square  root  is2x  —  Sy  —  4:Z. 

Is  every  polynomial  of  six  terms  the  square  of  a  trinomial? 
In  order  to  be  such  a  square,  how  many  squared  terms  must 
there  be  ?  What  is  the  sign  of  each  squared  term  and  why  ? 
How  are  the  signs  of  the  square  roots  of  these  squared  terms 
determined  ? 

EXERCISES 

In  each  of  the  following  determine  whether  the  polynomial 
is  a  perfect  square  and  if  so  find  its  square  root : 

1.  x^-{-y^-i-z^-2xy-]-2xz-2yz. 

2.  a'  -  S  ab  +  16b'  -2  ac  -h  c'  +  Sbc. 

3.  9a:^-\-4:y^-{-z^-12xy-^6xz-^yz. 

4.  y^-4:y^-'Sxf  +  16x-\-16a^  +  4:. 

5.  a'-^a'b'' -2  a'b  +  2  abc-  ab'c  +  b'c\ 

6.  a^-4a^  +  4a;^4-6a;3-12a;2-|_9. 

7.  x^  +  16  xy  +  2S9  +  S  x'^y  -\-34:X-{-lS6xy, 

8.  9a^  +  4a2  +  256-12a3-96a2+64a. 

9.  lQx^  +  4.y^-\-l-16x^y-{-Sx^-4.y. 

10.  25  +  49a^  +  4a;^-70a;-20a;2  +  28a^. 

11.  a'¥-2  a'b'  +  a"  -2  a'b^  -\-2  ab  +  b\ 

Besides  the  methods  of  factoring  which  have  been  applied  to 
the  types  of  expressions  thus  far  considered,  there  are  various 
other  processes  which  will  be  considered  in  the  Advanced  Course. 


SUMMAKY  OF  FACTORING  161 

SUMMARY  OF  FACTORING 

1.  Tell  how  to  decide  whether  a  polynomial  has  a  monomial 
factor.  Give  an  example  of  such  a  polynomial  and  find  its 
factors. 

2.  Tell  how  to  decide  whether  a  binomial  is  the  difference 
of  two  squares.     Give  such  a  binomial  and  find  its  factors. 

3.  Tell  how  to  decide  whether  a  trinomial  is  a  perfect 
square.  Distinguish  two  kinds.  Give  a  trinomial  square  of 
each  kind  and  factor  it. 

4.  Give  a  rule  for  factoring  the  sum  of  two  cubes. 

5.  Give  a  rule  for  factoring  the  difference  of  two  cubes. 

6.  Tell  how  to  decide  whether  a  trinomial  of  the  form 
x^ -\- bx -\- c  is  the  product  of  two  binomials. 

7.  Tell  how  to  decide  whether  a  trinomial  of  the  form  ax^  -\- 
6a;  -h  c  is  the  product  of  two  binomials. 

8.  Tell  how  to  factor  a  polynomial  by  grouping.  Give  a 
polynomial  which  can  be  factored  in  this  way  and  find  its 
factors. 

9.  Describe  the  square  of  a  trinomial.  Tell  how  to  decide 
whether  a  polynomial  is  a  perfect  square. 

MISCELLANEOUS  EXERCISES 

Classify  the  following  expressions  according  to  the  fore- 
going types  for  factoring,  and  find  the  factors : 


1. 

a^  +  5a;-f-6. 

7. 

2n^-6nc-Sny-\-9cy. 

2. 

1-Q^. 

8. 

4ar^-2/' 

3. 

a;a_|.iia._^30. 

9. 

a'  +  b\ 

4. 

4  cc^  +  9  /  + 12  «y. 

10. 

9  a^ -\- y* -{- 6  xf. 

5. 

4  or^  +  9  ?/"  —  12  »?/. 

11. 

2fa^-^4.ya'-Sya, 

6. 

6  ar^  +  4  ax  +  7  a;?/. 

12. 

x^  +  7x-\-6. 

162 


PRODUCTS,    QUOTIENTS,   AND  FACTORS 


13.  9x--^36y*-\-36xf. 

14.  9y  —  9z  —  2xy-\'2xz. 

15.  a^-l. 

16.  a*-i-b*-\-2a'b\ 

17.  a* -25. 

18.  27  a^- 125. 

19.  4:a^-\-4.ab  +  ab\ 

20.  4.a^-{-9x^-12ax\ 

21.  1  +  a^. 

22.  2a^  +  5ic  +  3. 

23.  36  +  4a^4-24»3. 

24.  (a; -1)2 -(a; +  1)2. 

25.  8  +  64a«. 

26.  ac  — aa;  — 4  6c  +  4  6a;. 

27.  27-216a3. 

28.  38+63a3. 

29.  25(a;  +  l)2-4. 

30.  5cx  —  10c-\-4:dx  —  Sd. 

31.  4(a;  +  2)2  +  2/2H-4(a;  +  2)y. 

32.  ra -h  2  rA  —  5  sa  — 10  s7i. 

33.  -2a'b  +  a*  +  b'. 

34.  2^a  — M+6a— 3  6. 

35.  3(a+l)34-4(a+l)2+a+l. 

36.  (a;  +  a)2  -  (a;  -  a)2. 

37.  15  m2  4- 224  m -15. 

38.  3iB2  +  27a;  +  42. 

39.  a;*  +  49a2  +  14aa^. 

40.  27tt«-a3a^. 


ce. 


41.  33a«  +  aV. 

42.  86d-406e  +  3cd-15 

43.  a;2-lla;+30. 

44.  (a;  +  2)2-4(a;-2)l 

45.  x*  +  9y^  —  6yx\ 

46.  4.a^-7  ca--4.d^  +  7cd\ 

47.  a2  +  15a-16. 

48.  18-27c  +  166-24c6. 

49.  4:-(a'  +  b^-2aby. 

50.  10r-{-Sbs  —  6br  —  5s. 

51.  25  +  64aj6  +  80aj3. 

52.  1000 -a^. 

53.  103  +  a;3_ 

54.  8ci3_|_a363  4.52^2^ 

55.  100 -49  a;*. 

56.  100  +  625  +  500. 

57.  a2-17a  +  72. 

58.  a^^  17a +  72. 

59.  a'-{-16b^-Sab. 

60.  x^  —  y^. 

61.  24a2  +  37a-72. 

62.  a;*  +  15aj2-100. 

63.  9«  +  8l 

64.  9a;^  +  16i/2  +  24a^2/. 

65.  1-1000. 

66.  16a'b^  +  24.ab-\-S6b\ 

67.  64  +  8. 

68.  16  a262  + 9  aV  + 24  a^ftc. 


MISCELLANEOUS  EXERCISES  163 

69.  a^-{-4:b^-^4.ab-4:x\  91.    a^-\-Sa' -ISO. 

70.  a86«+c^.  92.   a*-3a^-lS0. 

71.  5a^^lOfl^2/'4-30a^y.         93.   144-(a*4-6'-2a25). 

72.  16aV  +  4c2a;2_,_16^c2a;.      94.    81  a^ft^  +  49  c^  - 126  aft^c. 

73.  ay-^.  95.    12s2-23s^  +  10«2. 

74.  a;*-7a;2-120.  96.   36 x*  +  12 x'y*  +  f. 

75.  9a^62-12a'^6  +  4a2.  97.   16a:«-f  9  2/^4-24  a^^- 49. 

76.  8  a6H- 27  a6^  98.  /  + 35  2/ +  300. 

77.  a;^  +  4ar^  +  4-a^.  99.   5 y^ _ 80 y  +  300. 

78.  l-125a%\  100.   39a^-16a^4-l. 

79.  16  4-16a6  4-4a26^  101.   ac  -  be -^  ad  —  bd, 

80.  64a3  +  8a''68.  102.   625  -  (31- 4  a^^ 

81.  65r2  +  8r-l.  10S.*i^-\-ya-f!^-ay*, 

82.  a2-13a-140.  104.   x^ -\- 2  x^ -\- 1  -  a?, 

83.  a^  +  17a^  +  30.  105.    60 x^ -{- 7 xy-y^. 

84.  25-(a*-2a263  4-60.  106.    a^-20xy-^75yK 

85.  36a2_29a6  +  562.  107.   o.-^  -  17  a;  -  60. 

86.  a2-a-380.  108.   S6 a^b*  +  c^b* -\- 12 ab*c. 

87.  24  aV  +  a«  + 144  c»a2.  109.   4a2+9  6*+12a62_16a^ 

88.  9aj2H-42/^-12a^2_ig  ^^^    100 -(16ar^  +  /- 80^2/'). 

89.  81 4- 100 a^- 180 a^.  111.   6rc^-15re+22cd-55ce. 

90.  a*  +  27a2  +  i80.  112.    -  112aV  +  49a*  +  64c«. 

113.  4ar^  +  9/  +  22_12a;i/-4a^  +  62/;2. 

114.  d'b^  +  a"(^  +  b^c"-  2  a'bc  +  2  acb^  -  2  ab(^. 

115.  4a2-12a6  +  4ac+962  +  c2-66c. 

116.  a«-f  10a3  +  9a2  +  25-6a*-30a. 


164  PRODUCTS,   QUOTIENTS,   AND  FACTORS 

REVIEW  QUESTIONS 

1.  Define  a  positive  integral  exponent.  Explain  the  dit 
ference  between  an  exponent  and  a  coefficient. 

2.  Show  how  the  multiplication  of  monomials  may  depend 
upon  Principle  XIII. 

3.  Under  what  circumstances  are  exponents  added  in  mul- 
tiplication ?  State  Principle  XVI.  Use  this  principle  to  show 
that  (a^y  =  a\  (a^  =  a}'. 

4.  Under  what  circumstances  are  exponents  subtracted  in 
division?     State  Principle  XVII. 

5.  Show  how  the  division  of  monomials  may  depend  upon 
Principles  XV  and  XVII. 

6.  What  is  meant  by  factoring  ?  Is  the  following  expres- 
sion factored  ?     x  (a  -{-  b)  -{-  y  (a  -\-  b).     Why  ? 

7.  What  are  the  characteristics  of  a  trinomial  square  ? 
Are  the  following  trinomials  squares  ?     If  not,  state  what 

is  lacking.     x-\-xy  +  y^;  x'^ -^xh/-{-y'^j  a^  — 2a&  — 6^;  4a^-f- 
4  a6  4-  4  b\ 

8.  What  are  the  factors  of  the  difference  of  two  squares  ? 
Factor  x^  —  y^  as  the  difference  of  two  squares. 

9.  What  are  the  factors  of  the  sum  of  two  cubes  ? 
Factor  x^  -f-  y^  as  the  sum  of  two  cubes. 

10.  What  are  factors  of  the  difference  of  two  cubes  ? 
Factor  a^  —  y^  3is  the  difference  of  two  cubes. 

11.  Explain  how  to  factor  a  trinomial  by  inspecting  the 
end  products  and  cross  products  of  two  binomials. 

12.  By  means  of  the  following  examples  explain  the  process 
of  factoring  by  grouping. 

x^ -\- ax -\- bx  +  ab  j  a^  —  X —S  x^ -\- S. 

13.  What  are  the  characteristics  of  the  square  of  a  tri- 
nomial ?     Of  the  square  of  any  polynomial  ? 

14.  State  Principles  XVI  and  XVII  as  formulas  and  add 
them  to  your  list. 


DRILL  EXERCISES  165 

'  DRILL  EXERCISES 

1.  Eind  the  average  of  the  following  temperatures :  7  a.m., 
-4°;  8  a.m.,  -2°;  9  a.m.,  -1°;  10  a.m..  +1°;  11  a.m., +5°; 
12  m.,  +7°. 

2.J7a;H-(8aj  +  4)-5-2  =  4a;  +  9. 

3.  6a;  +  4(4a;4-2)=85-3(2a;-|t7). 

4.  8  +  7(6  +  6w)  +  2n  =  2(4n-f5)+18n  +  49. 

5.  5(9aj  +  3)+6a;  =  24a;-4(3a;  +  2)  +  36. 

6.  ^a^^  +  ^)+13  +  5x-6  =  47. 

4 

„      a  —  h  4:C  a  —  b 


2x-l     (2a;  +  l)(2a;-l)     2a;  +  l 
x  +  2     x-1  ,  3a;4-2 

o.    = 1 • 

a  b  db 

l5a;  +  82/  =  l.  '    \2x  +  y  =  l. 

1 2a; -f  3  2/ =  314  + 13  j^.  "    l3'a;-2^  =  3. 

U-22^  =  42^  +  3.  *    l2a;H-2-62/  =  2-a;. 

ra;  +  22/  +  2z  =  3,  |'2a;  +  52/  +  72  =  7, 

i  3a;-42/  +  2  =  19, 


15.    "I  3a;-42/  +  2  =  19,  16.    j  3fl;- 9y- 2z  =  23, 

t  _  2 a;  +  6  ?/  +  32  =  0.  •  t  -  a;  +  3y  +  3  z=  -10. 

17.    Graph  the  equations :  x  —  ^y  =  l  and  2  a;  +  ?/  =  4. 

18.'  The  older  of  two  sisters  is  now  8  years  less  than  twice  as 
old  as  the  other.  If  x  represents  the  age  of  the  younger  sister, 
represent  in  symbols  twice  the  sum  of  their  ages  7  years  ago. 

19.  A  rear  wheel  of  a  wagon  has  a  circumference  4  feet 
greater  than  that  of  a  front  wheel.  If  x  represents  the  num- 
ber of  feet  in  the  circumference  of  the  rear  wheel,  represent  in 
symbols  the  number  of  revolutions  each  wheel  must  make  to 
go  one  mile. 


CHAPTER   X 
EQUATIONS  SOLVED  BY  FACTORING 

160.  Illustrative  Problem,  There  are  two  consecutive  integers 
the  sum  of  whose  squares  is  61.     What  are  the  numbers  ? 

Solution.     It  X  =  one  of  the  numbers,  then  a;  +  1  is  the  other. 

Hence,  a;^  +  (x  +  1)2  =  61  (1) 

By  F,  a;2  +  a;2  +  2a:  +  l  =  61  (2) 

ByF,I,S,  2a:2+2a;  =  60  (3) 

By  Z>,  x^  +  x  =  30  (4) 

Equation  (4)  differs  from  any  which  we  have  studied  here- 
tofore in  that  it  contains  the  square  of  the  unknown  number, 
after  all  possible  reductions  have  been  made. 

161.  Definition.  Equations  which  involve  the  second  but  no 
higher  degree  of  the  unknown  number  are  called  quadratic 
equations. 

One  method  of  solving  quadratic  equations  is  now  to  be 
considered. 

By  Sj  equation  (4)  above  may  be  written 

•a;*2  +  a:  _  30  =  0.  (5) 

Factoring  the  left  member, 

(a:  +  6)(x-5)=0.  (6) 

This  equation  is  satisfied  by  a;  =  5  since  (5  +  6)(5  —  5)  =  11  •  0  =  0, 
and  also  by  a:  =  6  since  (- 6  +  6)(- 6  -  5)  =  0  •  (- 11)=  0. 

Test  by  substitution  whether  5  and  5+1  =  6  satisfy  the  conditions 
of  the  problem ;  also  whether  —  6  and  —  6  +  1  =  —  5  satisfy  it. 

It  thus  appears  that  equation  (4)  has  two  roots,  namely,  5  and  —  6. 
The  two  pairs  of  corresponding  integers  5,  6  and  —6,-5  each  satisfy 
the  conditions  of  the  problem. 

166 


EQUATIONS   SOLVED   BY  FACTORING  167 

EXERCISES 

1.  If  one  of  two  factors  is  zero,  what  is  the  product  ?  Does 
it  matter  what  the  other  factor  is  ? 

2.  Find  a  value  of  q?  which  makes  {x  —  3)(x-\-2)  equal  to 
zero.  Does  this  value  of  x  make  both  factors  equal  to  zero? 
Is  it  necessary  that  both  factors  should  be  made  equal  to  zero  ? 

3.  Find  a  value  of  x  which  satisfies  the  equation 

(x-7)(x^-\-2x^3)=0', 
also  one  which  satisfies  (x  -f-  8)(ar*  -|-  a;  +  4)  =  0. 

Suggestion.  Find  a  value  of  x  which  makes  the  first  factor  zero 
in  each  case. 

4.  Find  two  values  of  x  which  satisfy  (x  —  3)  (a:  -|-  4)  =  0, 
also  two  which  satisfy  (x  +  8)(a;  —  3)  =  0. 

5.  Find  two  values  of  x  which  satisfy  5  ic(a;  4- 7)  =  0.  Does 
x=  0  satisfy  this  equation ? 

6.  Find  two  values  of  x  which  satisfy  (3  a;  —  2)  (2  a;  +  5)  =  0. 

162.  The  method  of  solution  suggested  by  the  foregoing  ex- 
amples consists  of  three  steps  : 

(1)  Transform  the  equation  so  that  all  terms  are  collected  in 
one  member,  with  similar  terms  united^  leaving  tJt^  other  member 
zero.  This  can  always  be  done  by  Principle  VI.  It  is  con- 
venient to  make  the  right  member  zero. 

(2)  Factor  the  expression  on  the  left. 

(3)  Find  the  values  of  x  which  make  each  of  these  factors  zero. 
This  is  readily  done  by  setting  each  factor  equal  to  zero  and  solv- 
ing it  for  the  unknown. 

EXERCISES 

Find  two  solutions  for  each  of  the  following  quadratic  equa- 
tions : 

1.  a^-3a;  +  2  =  0.  5.    a2  4-10a-f8=-3  a-34. 

2.  ar^  +  7a;  =  30.  6.   a^H-S  a  =  10  a  + 18. 

3.  a2-lla  =  -30.  7.    a^  +  lO  a  =  - 24-4  a. 

4.  a2  +  13a  =  30.  8.   2  ar^  -  6  a;  =  -  40 -|- 12  a?. 


168  EQUATIONS  SOLVED  BY  FACTORING 

9.   Sx-{-x'  =  20x-72.  12.   11  a;  +  3  ar^  =  20. 

10.  17a;  +  30  =  ~aj2_40.  13.   16-5x-\-x'  =  -2  x"- 

11.  7x^-{-2x  =  30x-21.  20x-2. 

14.  0^-16=0.     15.  0(^-1=0.     16.  x?-x=0.     17.  a^+aj=0. 

18.  4:x'  =  25.      19.   a;2^43._,_4^Q       go    a^^^S  a;  +  16  =  0. 

21.  a^  +  12a;  +  6  =  5ic-4.        23.    60  a;+'4  a;2  4-144  =  8  ic. 

22.  2aj2-7a;  =  60  +  7a;.  24.    180^  =  63-0;^. 

25.  24:0^=12 x-\-12.     27.  22a;+a^=363.     29.  2a;2=2-3a;. 

26.  2a;=63-a;2.  28.  3x^-\-7x  =  6.       30.  a;-2=-3a^. 

163.  It  is  sometimes  possible  by  the  above  process  to  solve 
equations  in  which,  the  exponent  of  the  highest  power  of  the  un- 
known is  greater  than  2. 

Ex.  1.     Solve  the  equation : 

a:3  +  30  ar  =  11  x^.  (1) 

By  S,  x^-llx-^  +  dOx=  0.  (2) 

By  §  141,  x(x^  -  11  X  +  30)  =  0.  (3) 

By  §  154,  x(x  -5)(x-6)=  0.  (4) 

(4)  is  satisfied  if  a:  =  0,  if  x  —  5  =  0,  and  if  a;  —  6  =  0. 

Hence  the  roots  are  a:  =  0,  a:  =  5,  a:  =  6. 

Ex.  2:    x(x  V  l)(x  -  2)(«  +  3)  =  0. 

Any  value  of  x  which  makes  one  of  these  factors  zero  re- 
duces the  product  to  zero  and  hence  satisfies  the  equation. 
Hence  the  roots  of  the  equation  are  found  from 

a;  =  0,  a;-f-l  =  0,  a;-2  =  0,  anda;4-3  =  0. 

That  is,  a?  =  0,  a^  =  —  1,  a;  =  2,  a;  =  —  3  are  the  values  of  x 
which  satisfy  the  equation. 

Notice  that  this  process  is  applicable  only  when  one  member 
of  the  equation  is  zero  and  the  other  member  is  factored. 

Solve  the  following  equations  by  factoring: 

1.  x^-x^  =  6x.  3.   ar^-25a;  =  0. 

2.  6a;  =  4iB2  +  a^.  4.   a^  -  3ar^  =  -  2a;. 


PROBLEMS   SOLVED  BY  FACTORING  169 

5.  3  a;  3  =  15  ar^  4- 42  a;.  10.  a?  —  ax -^hx  —  ah  =0. 

6.  6  a:^  +  315  aj  =  80  x\  11.  o?  +  ax —  hx  —  ah  =  0. 

7.  a^  +  aa;  +  6a;  +  a6  =  0.  12.  9  (a;  +  2/  -  4  (a;  -  3)^  =  0. 
S.  x^  —  ax  —  hx  +  ah  =  0.  13.  a^  —  a;  —  3  a^  +  3  =  0. 

9.    4(a;- 2)2 -(a; +  3)2  =  0.      14.   a^- 4x  -  8  a^  +  32  =  0. 

PROBLEMS  SOLVED  BY  FACTORING 

164.  Illustrative  Problem.  The  paving  of  a  square  court 
costs  40  /  per  square  yard  and  the  fence  around  it  costs  $  1.50 
per  linear  yard.  If  the  total  cost  of  the  pavement  and  the 
fence  is  $  100,  what  is  the  size  of  the  court  ? 

Solution.     Let  x  =  the  length  of  one  side  in  yards. 
Then  40  x^  =  cost  in  cents  of  paving  the  court, 

and  150  •  4  x  =  600  x  =  cost  of  the  fence  in  cents. 

Hence,  40  x^  +  600  a:  =  10000.  (1) 

By  i>,  x^+16x=  250.  (2) 

By  5,  x^  +  15x  -  250  =  0.  (3) 

Factoring,  (x  -  10)(a:  +  25)  =  0.  (4; 

Whence,  x  =  10,  and  also  x  =  —  25.  (5) 

It  is  clear  that  the  length  of  a  side  of  the  court  cannot  be  —  25 
yards.  Hence  10  is  the  only  one  of  these  two  results  which  has  a 
meaning  in  this  problem. 

It  happens  frequently  when  a  quadratic  equation  is  used  to 
solve  a  problem  that  one  of  the  two  numbers  which  satisfy  this 
equation  will  not  satisfy  the  conditions  of  the  problem. 

PROBLEMS 

In  each  of  the  following  problems  find  all  the  solutions  pos- 
sible for  the  equations  and  then  determine  whether  or  not  each 
solution  has  a  reasonable  interpretation  in  the  problem. 

1.  The  dimensions  of  a  picture  inside  the  frame  are  12  by  16 
inches.  What  is  the  width  of  the  frame  if  its  area  is  288 
square  inches  ? 


170 


EQUATIONS   SOLVED   BY  FACTORING 


37 

si 

x-iQ       1^5 : 

s| 

Hi 

^1 

! 

x-lO        I 

.± 

P^ 

160 


80 


a;(160-2x) 
160 -2x 


2.    An  open  box  is  made  from  a  square  piece  of  tin  by  cutting 
out  a  5-inch  square  from  each  corner  and  turning  up  the  sides. 
How  large  is  the  original   square  if  the  box 
contains  180  cubic  inches? 

If  a;  =  length  of  a  side  of  the  tin,  then  the  volume  of 
the  box  is:  5(a;-10)(a:  -  10)  =  180.   (See  the  figm-e.) 

3.  A  rectangular  piece  of  tin  is  4  inches 
longer  than  it  is  wide.  An  open  box  contain- 
ing 840  cubic  inches  is  made  by  cutting  a  6-inch  square  from 
each  corner  and  turning  up  the  ends  and  sides.  What  are 
the  dimensions  of  the  box? 

4.  A  farmer  has  a  rectangular  wheat  field  160  rods  long  by 
80  rods  wide.  In  cutting  the  grain,  he  cuts  a  strip  of  equal 
width  around  the  field. 
How  many  acres  has 
he  cut  when  the  width 
of  the  strip  is  8  rods  ? 

5.  How  wide  is  the 
strip  around  the  field 
of  problem  5,  if  it  con- 
tains 27i  acres  ? 

6.  In  the  northwest  a  farmer  using  a  steam  plow  starts  plow- 
ing around  a  rectangular  field  640  by  320  rods.  If  the  strip 
plowed  the  first  day  lacks  16  square  rods  of  being  24  acres,  how 
wide  is  it  ? 

7.  A  rectangular  piece  of  ground  840  by  640  feet  is  divided 
into  4  city  blocks  by  two  streets  60  feet  wide  running  through 

840  feet  it  at  right  angles.     How  many  square  feet 

are  contained  in  the  streets  ?' 

8.    A  farmer  lays  out  two  roads  through 
the  middle  of  his  farm,  one  running  length- 
wise of  the  farm  and  the  other  crosswise. 
How  wide  are  the  roads  if  the  farm  is  320 
by  240  rods,  and  the  area  of  the  roads  is  1671  square  rods  ? 


160-2a? 


QUADRATIC   AND   LINEAR   EQUATIONS  171 

QUADRATIC  AND  LINEAR  EQUATIONS 
165.    When  two  simultaneous  equations  are  given,  one  quad- 
ratic and  one  linear,  they  may  be  solved  by  the  process  of  sub- 
stitution, which  was  used  (§  122)  in  the  case  of  two  linear 
equations. 

Illustrative  Example.     Solve  the  equations : 

(^-f  =  -m  (1) 

[x-Sy  =  -12.  (2) 

From  (2)hj  S,  x  =  Zy  -  12.  (3) 

Substituting  (3)  in  (1),     (3  y  -  12)2  _  3,2^  _  16.  (4) 

From  (4)  by  i^,         9  3^2  _  72  y  +  144  -yj^  =  -  16.  (5) 

From  (5)  by  F,  ^,  8  3/2  _  70  3,  +  160  =  0.  (6) 

By  A  2/2  _9y  + 20  =  0.  (7) 

Factoring,  {y  -  5)  (y  -  4)  =  0.  (8) 

Hence,  y  =  5,  and  y  =  i.  (9) 

Substitute  y  =  5  in  (3)  and  find  x  =  S. 

Substitute  y  =  4  in  (3)  and  find  x  =  0. 

Therefore  (1)  and  (3)  are  satisfied  by  the  two  pairs  of  values, 

X  =  3,  y  =  5,  and  x  =  0,  y  =  4. 
Check  by  substituting  these  pairs  of  values  in  (1)  and.  (2). 

EXERCISES 

In  the  manner  just  illustrated  solve  the  following : 


2 


x  +  2y  =  S,  (x-2y  = 

6a;2H-12/  =  128.  '     [x'-6y'  =  10. 

(x  +  y=l,  (Sx=16y  =  -120, 

[x'-{.y^  =  l.       ^  '    l7ar^-f  2/  =  585. 

(2x-y  =  6,  (Tx-{-9y  =  SS, 

(x-{-3y  =  6,  (x-y  =  6, 

10:24.32,2^12.  .                     ■     U'2-72/2  =  36.     . 


172  EQUATIONS  SOLVED  BY  FACTORING 


9. 


_,  w  =  —  11,  f  x  —  9y  =  2, 

10.       „.  .        „   .       '  16.     '  -^ 


162/^=11.  U2-452/2  =  4. 


11. 


12. 


x  —  y  =  —  7,  ( x-{-y  =  Sj 

Ua^  +  32/2  =  147.  ^'^'    1 13 0^2 +  3 2/2  =  160. 

(x-y==2,  (2x-5y  =  -16, 

[x^-5y^  =  4:.  '    [Ax' -{-15  y' ^256. 


fa;  —  ^  =  1,  r7a;  +  4v  =  7, 

13.    i  19      ' 


2/2  =  -5.  I49a;2_8^^49^ 

r5.'c-72/  =  -28.  fa; -32/ =  -12, 

^^*    ll5a'2  +  49/  =  784.  ^^'    I  a;^  -  /  =  - 16. 

166.  If  squares  are  constructed  on  the  two  sides,  and  also  on 
the  hypotenuse  of  a  right-angled  triangle,  then  the  su7n  of  the 
squares  on  the  sides  is  equal  to  the  square  on  the  hypotenuse. 
This  is  proved  in  geometry,  but  may  be  verified  by  counting 
squares  in  the  accompanying  figure.  This  proposition  was  first 
discovered  by  the  great  philosopher  and  mathematician  Py- 
thagoras, who  lived  about  550  e.g.  Hence  it  is  called  the 
Pythagorean  proposition. 

We  now  proceed  to  solve  some  problems  by  this  proposition. 

PROBLEMS 

1.  The  sum  of  the  sides  about  the  right  angle  of  a  right  tri- 
angle is  35  inches,  and  the  hypotenuse  is  25  inches.  Find  the 
sides  of  the  triangle. 

Solution.     Let  a   =  the  length  of  one  side  in  inches, 
and  h   =  the  length  of  the  other. 

Then  a  +  b   =35,  (1) 

and  a^  +  &2  _  252  =  625      (Pythagorean  proposition).         (2) 

From  (1),  a  =  35 -J. 


QUADRATIC   AND  LINEAR  EQUATIONS  173 




%^\- 

jL    ^\ 

t 

it      s 

■"     _  _    :^__ 

It  _::5^__: 

it        ^s              ~"  " 

'                    s^ 

I 

It      :        V" 

:                            z        -j- 

It      -  it  -S 

It      __      ---^^ : 

—    __    __      __ ^  -- 

i               f5  '  .n    f  t                  ^  s 

^ 

it      :              :  :^w_: 

4:                            ^y 

__  ..J.    __    :  _ 

__^_ 

»                                                                          w 

A 

^   1'-t~                     ^ 

^  _ 

1  q.  ru  1                                y 

1:.,:::1:_    __        :^:_: 

1             R  '— 1     f  t            i 

^ 

:    :  : s~          _    _ 

_it:_      _  .A.    __ 

•*  ^                         Re 

77f       II                  1 

:__:_:_      _  _,:___:  ^»,^±:  ^!j 

f:±__±___  :_/  :: 

5^ 

--It            IL    -    _X 

-t            % 

_     -     B&i..ft.             -                            -  h 

i         ~t--       -- 

!r                                     5  ^  - 

it         > 

::_:_:        _  __:       i.^; 

IT    _J_ 

si — i  i^i       I — 

^^     it   t 

-^s  i> 

<Xi- 

Ji 

Substitutiug  in  (2),  (35  -  Z;)"^  +  ft^^  625, 

or  1225  -  70  6  +  62  +  &2  ==  625, 

2  &2  _  70  6  +  600  =  0, 
fta  _  35  &  +  300  =  0, 

Factoring,  (5  -  20)  (6  -  15)  =  0. 

Whence  h  -  20,  and  h  -  15. 

From  (1),  if  &  =  20,  a  =  15,  and  if  6  =  15,  a  =20;  that  is,  the 
sides  of  the  triangle  are  15  and  20. 

2.  The  difference  between  the  two  sides  of  a  right  triangle 
is  2  feet,  and  the  length  of  the  hypotenuse  is  10  feet.  Find 
the  two  sides. 


174  EQUATIONS   SOLVED   BY   FACTORING 

3.  The  sum  of  the  length  and  width  of  a  rectangle  is 
17  rods,  and  the  diagonal  is  13  rods.  Find  the  dimensions  of 
the  rectangle. 

4.  A  room  is  3  feet  longer  than  it  is  wide,  and  the  length 
of  the  diagonal  is  15  feet.     Find  the  dimensions  of  the  room. 

5.  The  length  of  the  molding  around  a  rectangular  room  is  46 
feet,  and  the  diagonal  of  the  room  is  17  feet.  Find  its  dimensions. 

6.  The  longest  rod  that  can  be  placed  flat  on  the  bottom 
of  a  certain  trunk  is  45  inches.  The  trunk  is  9  inches 
longer  than  it  is  wide.     What  are  its  dimensions  ? 

7.  The  floor  space  of  a  rectangular  room  is  180  square  feet, 
and  the  length  of  the  molding  around  the  room  is  56  feet. 
What  are  the  dimensions  of  the  room  ? 

8.  A  rectangular  field  is  20  rods  longer  than  it  is  wide,  and 
its  area  is  2400  square  rods.     What  are  its  dimensions  ? 

9.  A  ceiling  requires  24  square  yards  of  paper,  and  the  border 
is  20  yards  long.     What  are  the  dimensions  of  the  ceiling  ? 

10.  The  area  of  a  triangle  is  18  square  inches,  and  the  sum 
of  the  base  and  altitude  is  12.     Find  the  base  and  altitude. 

11.  The  altitude  of  a  triangle  is  7  inches  less  than  the  base, 
and  the  area  is  130  inches.     Find  the  base  and  altitude. 

12.  The  sum  of  two  numbers  is  17,  and  the  sum  of  their 
squares  is  145.     Find  the  numbers. 

13.  The  difference  of  two  numbers  is  8,  and  the  sum  of  their 
squares  is  274.     Find  the  numbers. 

14.  The  difference  of  two  numbers  is  13,  and  the  difference 
of  their  squares  is  481.     Find  the  numbers. 

15.  The  sum  of  two  numbers  is  40,  and  the  difference  of 
their  squares  is  320.     Find  the  numbers. 

16.  The  sum  of   two  numbers  is  45,  and  their  product  is 
450.     Find  the  numbers. 

17.  The  difference  of  two  numbers  is  32,  and  their  product 
is  833.     What  are  the  numbers  ? 


DRILL   EXERCISES 
DRILL  EXERCISES 


175 


6  2 

2.   15  I  21(3 +  r.)  ^  2(6  +  18ar)_3(9^  +  12)  ^  gg 
/  o  o 

,     ll(5a;  +  25)   ,  3(6 a; -  2)  _ 7(4  a;  +  8)  ,  12a;  +  36  . 

3. 1  ^r  —  -  I -f- 


35. 


4    2(3:4-1)      3^      0^-1 

X  —1  35  +  1 


5.    £±^=5 


6. 


7. 


8. 


a;  — a 
5a;-3     2y 


4(a;  —  g) 
x-\-a 


h 


3  a;  4- 5      y- 
2       "^'    5 


10 


6. 


2a;-9     4y-2^ 
3  5      ~ 

7a;4-4     y  +  6_o 


1, 


.3  6" 

4a  +  7    76  +  3. 


1 

2' 

=  -5. 


9. 


10. 


11. 


2a4-5     36-lO^Q 


3a4.6      5-46 

8  7 


3      4^6        * 

2     8     12       ' 

?--^  +  -^  =  l. 
6     2     3 


x  +  y-z  =  S5y 

?4.^  +  ,  =  15, 
3^5^ 

1  +  1-^  =  3. 
5      5 


=  2. 


12.  If  r  represents  the  rate  in  miles  per  hour  at  which  a 
train  is  moving,  how  far  will  it  go  in  t  hours  ?  Another  train 
runs  10  miles  per  hour  faster.  Express  in  symbols  the  sum  of 
the  distances  which  these  two  trains  travel  in  t  hours. 

13.  If  ?*i  represents  the  rate  at  which  a  river  is  flowing  and 
ra  the  rate  at  which  a  steamer  can  go  in  still  water,  express  in 
symbols  the  distance  which  the  steamer  can  go  in  t  hours :  (a) 
down  the  river ;  (b)  up  the  river. 


CHAPTER  XI 
SQUARE  ROOTS  AND  RADICALS 

167.  Definition.  A  square  root  of  a  number  is  one  of  its  two 
equal  factors. 

Thus  3  is  a  square  root  of  9,  since  3  •  3  =  9.  Similarly  a  +  b  h  a 
square  root  oi  a^  +  2ab  +  b'\ 

It  should  be  noted  that  every  square  has  two  square  roots. 
E.g.    —  3   is   also  a   square    root  of    9   as  well    as    +  3,    since 
(-3).  (-3):=  9. 

The  positive  square  root  of  a  number  is  indicated  by  the 
radical  sign  V  alone  or  preceded  by  the  sign  -}-.  The  nega- 
tive square  root  is  indicated  by  the  radical  sign  preceded  by 
the  sign  — . 

Eg.   +  >/9  or  VO  =  +  3  and  not  -  3,  and  -  \/9  =  -  3,  and  not  +  3. 

The  square  root  of  any  number  is  at  once  evident  if  we  can 
resolve  it  into  two  equal  groups  of  factors. 

E.g. 
V576  =  \/2.2.2.2.2.2.3.3  =  \/(23  •  8)  (2^ .  3)  =  \/2rr2i  =  24. 

EXERCISES 

Find  by  inspection  the  following  indicated  square  roots : 


1. 

V4. 

8. 

V121. 

15. 

V324. 

22. 

-V2\ 

2. 

V9. 

9. 

-  Vl69. 

16. 

V289. 

23. 

-  VB'K 

3r 

-  Vl6. 

10. 

■y/225. 

17. 

-V625. 

24. 

-  V7«. 

4. 

V25. 

11. 

Vl96. 

18. 

-V900. 

25. 

V^. 

5. 

V36. 

12. 

-V256. 

19. 

Vioooo. 

26. 

-  V3^^. 

6. 

-V49. 

13. 

-V576. 

20. 

v^. 

27. 

v^. 

7. 

VsT. 

14. 

V400. 

21. 

-V^. 

28. 

v^. 

176 


SQUARE   ROOTS  AND   RADICALS  177 

168.  The  square  root  of  the  product  of  several  factors,  each 
of  which  is  a  square,  may  be  found  in  two  ways  if  the  factors 
are  expressed  in  Arabic  figures. 


E.g,  V4  .  16  .  25  =  V1600  =  ViO  •  40  =  40, 


But  with  literal  factors,  the  second  process  only  is  available. 


E.g.  VWcfib^  =  \/42a2(/;2)2c2  =  ^ab^c. 
EXERCISES 

Find  the  following  indicated  square  roots : 


1. 

-V2=^-3«. 

7. 

8. 

9. 
10. 
11. 
12. 

-V3^.5". 

13. 
14. 
15. 
16. 
17. 
18. 

V9.<y^. 

2. 

V81  .  121. 

-V222.3i^. 
Vl6a'b'c\ 
V64aV. 
-V4^a26*. 

-V121aV. 

3. 
4. 

V49  .  25  .  169. 
-V82.  5=^.31 

V625ar*?/2 

5. 

V5* .  32 .  4*. 
V25  .  36. 

V1225a^^ 

6. 

-V36  6^c*". 

Notice  that  V9  +  16  is  not  equal  to  V9  +  VI6. 
The  preceding  exercises  illustrate 

Principle  XVIII 

169.  Rule.  The  square  root  of  a  product  is  obtained 
by  finding  tlie  square  root  of  each  factor  separately  and 
then  taking  the  product  of  tliese  roots.    TJiat  is, 

Va  •  b  =  Va  •  V6- 

In  order  that  a  factor  may  be  a  perfect  square  it  must  be 
a  power  whose  exponent  is  even.  Its  square  root  is  then  a 
power  of  the  same  base  whose  exponent  is  equal  to  one-half 
the  given  exponent. 

Thus,  V^s  =  y/x^  '  x^  =  x^  =  x^^'^. 

Hence  to  find  the  square  root  of  a  monomial  we  divide  the 
exponent  of  each  factor  by  2. 


178  SQUARE   ROOTS  AND   RADICALS 

EXERCISES 

Find  the  following  square  roots  : 

1.  -V4aW.  6.    -VlO^"a^.  11.  V81  ^fc^\ 

2.  -V3Vyi  7.    V5*^  12.  V729ayVl 

3.  -^^W^WFK  8.    V5^ .  38 .  72.  13.  -V64-625a26^ 

4.  V12I  icyi  9.    V3".7^^a^  14.  V256«V^ 

5.  -V576^¥:         10.    -V25^^6W  15.  V3^T5^. 

SQUARE  ROOTS  OF  POLYNOMIALS 

170.  In  §  159  the  square  roots  of  certain  polynomials  were 
found  by  inspection.  Any  polynomial  square  may  be  recognized 
and  the  square  root  found  by  that  method,  provided  no  similar 
terms  have  been  combined  in  the  square. 

171.  The  case  where  some  terms  of  the  square  combine  is 
illustrated  by  the  example : 

(a:2  -t-a:+  \y  =  x^  +  2^2  +  1  +  2  a:3  +  2  x^  +  2  x^x^  +  S  xH  3a:2+2x+l. 

Study  the  answers  to  the  following  questions.  They  are 
needed  in  finding  the  square  root  of  such  an  expression : 

1.  What  terms  must  be  added  to  a?  to  make  it  the  square 
of  a  +  6  ? 

2.  What  terms  must  be  added  to  a^  +  2  a6  +  &^  to  make  it 
the  square  of  a  +  6  +  c  ? 

3.  What  terms  must  be  added  to  a^+2  db-\-h^+2  ac+2  hc-^c? 
to  make  it  the  square  of  a -\-h-\-c  +d. 

The  above  answers  may  be  summarized  as  follows : 

When  a  term  is  added  to  a  polynomial,  the  square  of  the  result- 
ing polynomial  is  equal  to  the  square  of  the  original,  plus  twice 
the  product  of  the  new  term  and  the  sum  of  the  original  terms, 
plus  the  square  of  the  new  term. 

This  is  expressed  by  the  following  formula :  (a  -f  6  +  c  +  </)* 


SQUARE   ROOTS  OF  POLYNOMIALS  179 

172.    Illustrative  Example.     Find  the  square  root  of 
16  a^  -  24  ar^  -f-  25  a;^  -  52  a:^  +  34  ic2  _  20  a;  -h  25. 

Solutio7i 
a      +  b     4-  c  +  rf 


Square  root           4  ar«  - 

-3a:2+2a:-5 

Given  square        16  a;« 
a2  =  16  a:« 

-  24  a:5  +  25  x*  -  52  x^  +  34  a:2  -  20  a;  +  25  (1) 

2  aft  4-  6*  = 

-  24  a:5  +  25  a;*  -  52  2:8  +  34  a;2  _  20  a;  +  25  (2) 
-24a:«+    9  a:*                                                      (3) 

2(a  +  h)c  +  c2  = 
2(a  +  b  +  c)d  +  rf2  = 

16x4  _  52x8  +  34^.2  _  20x  +  25  (4) 
16  X*  -  12  x8  +    4  x2                         (5) 

-  40x8  +  30x2  -  20x  +  25  (6) 

-  40x8  +  30x2 -20x  + 25  (7) 

Explanation.  First  arrange  the  terms  according  to  increasing  or 
decreasing  powers  of  the  letter  involved,  as  in  long  division.  The 
first  term  of  the  root  is  the  square  root  of  the  first  term  of  the  poly- 
nomial.    The  square  of  this  is  subtracted,  leaving  (2). 

The  next  term  of  the  root,  —  3  x2,  is  found  by  dividing  the  first 
term  of  (2)  by  2(4  x8)=  Sx^,  that  is,  -  24x6  ^8x8  zi: -3x2,  corre- 
sponding to  2  aft  -f-  2  a  =  6. 

We  then  s^ubtract  2(4  x8)(-  3  x2)  +  (  -  3  x2)2=  _  24  x^  +  9  x*,  corre- 
sponding to  2  aft  +  ft2,  and  this  completes  the  subtraction  of  (4  x8  -  3x2)2. 

Subtracting  (5)  completes  the  subtraction  of  (4x8  _  3  ^2  +  2  x)2, 
and  subtracting  (7)  completes  the  subtraction  of  (4  x8  — 3  x2-|-2  x  — 5)2. 

At  each  step  the  next  term  of  the  root  is  found  by  dividing  the 
first  term  of  the  remainder  by  twice  the  first  term  of  the  root.  Thus 
from  (4)  the  third  term  of  the  root  is  16  x*  h-  2(4x8)  =  2x,  corre- 
sponding to  2  ac  -^  2  a  =  c. 

Since  the  remainder  is  zero  after  the  last  subtraction,  this  shows 
that  (4  x8  -  3  x2  +  2  X  -  5)2  is  exactly  the  given  polynomial. 

EXERCISES 

Eind  the  square  roots  of  the  following  polynomials : 

1.  a^  +  2a^H-3ar^-f  2aj  +  l.      3-   1 -f  2  &  -  ft^- 2  ft^-j-ft^ 

2.  l-2a-f  3a2-2a^  +  a'.      4.    a*  +  4a^-f  6a24-4a  +  l. 


180 


SQUARE   ROOTS  AND   RADICALS 


5.  c^_.4c3  +  6c2-4c  +  l. 

6.  a;*  — 2a^  +  5a;2  — 4a;  +  4. 

8.  x^-Ax^y  +  6a^y^-4:xy^  +  y\ 

9.  «^  +  53a2  +  14a3  +  28a  +  4. 

10.  a^  +  6a^  +  15a*  +  20a3  +  15a2  4.6a  +  l. 

11.  a«-6a^  +  15a^-20a3  +  15a2-6a  +  l. 

12.  4a:*5-12a:^  +  13a;*-14a^-f  13a;2-4a;  +  4. 

13.  16a«  + 24a^ +  25  a*  +  20  a«  + 10  a2  +  4a  +  l. 

14.  a;yH-2ar'/  +  3a;y  +  4a;32^3_|.3^2^2^2a;y  +  l. 

15.  l  +  2x-{-3x^  +  4ta^  +  5x^-{-4:X^-{-Sa^  +  2x'^  +  a^. 


SQUARE  ROOTS   OP   NUMBERS   EXPRESSED   IN  ARABIC   PIOURES 
173.   The   square   root   of   a   number   expressed   in   Arabic 
figures  may  be  found  by  the  process  just  used  for  polynomials. 

Illustrative  Example.     Find  the  square  root  of  405769. 
Solution. 


Square  Root            Sqtjark 

a  +  6  +  c 

600  +  30  +  7  =  637       405769 

a2  =  600'       360000 

2a&  =  36000 

45769 

62  =   900 

36900      [ 

36900 

8869 

2(a  +  6)c  =  8820 

c2  =   49 

8869 

8869 

0 

Explanation.  We  see  that  the  given  number  is  greater  than  the 
square  of  600  and  less  than  the  square  of  700.  Hence  we  take  600  as 
the  first  term  of  the  root  and  subtract  600    from  the  number. 

As  in  the  case  of  the  polynomial  the  second  term  of  the  root  is 


SQUARE  ROOTS  OF  NUMBERS  EXPRESSED  IN  FIGURES     181 

found  by  dividing  the  remainder  by  twice  the  first  term  of  the 
root,  that  is  45769  -=- 1200.  This  gives  a  quotient  greater  than  30  and 
less  than  40  and  hence  30  is  the  next  term  of  the  root. 

The  third  term  of  the  root  is  found  by  dividing  8869  by  2  x  630, 
that  is,  by  2(a  +  b).  In  the  case  of  a  polynomial  it  is  sufficient  to 
divide  by  2  a,  but  with  arithmetical  numbers  the*third  term  is  usually 
more  easily  found  if  we  divide  by  2  (a  +  6). 

The  remaining  parts  corresponding  to  (a  +  by  and  (a  +  6  +  c)2  are 
now  computed  and  subtracted  exactly  as  in  the  case  of  a  polynomial. 

174.  The  first  term  of  the  root  may  be  found  as  follows : 
Separate  the  number  into  groups  of  two  digits  each  from  the 

decimal  point  toward  the  left.  Take  the  square  root  of  the  largest 
square  contained  in  the  last  group  to  the  left  and  adjoin  one  zero 
for  each  remaining  group. 

Thus  in  the  root  of  87  23  56  the  first  term  is  900,  since  81  is  the 
largest  square  in  the  left  group  and  there  are  two  other  groups. 
In  the  root  of  7  34  86.593,  the  first  term  is  200,  since  4  is  the  largest 
square  contained  in  the  left  group  and  there  are  two  other  groups 

EXERCISES 

Find  the  square  root  of  each  of  the  following : 

1.  294,849.         5.   3481.  9.    100,489.  13.  357.21. 

2.  37,636.  6.   7569.  10.     265.69.  14.  16,641. 

3.  872,356.         7.   1849.  11,    87.4225.  15.  32,761. 

4.  599,076.         8.   73,441.        12.    170,569.  16.  2332.89. 

175.  In  case  the  number  whose  square  root  is  to  be  found 
has  no  figure  to  the  left  of  the  decimal  point,  the  first  term  of 
the  root  may  be  found  by  the  following  rule : 

Separate  the  number  into  groups  of  two  digits  from  the  decimal 
point  toward  the  right.  If  necessary,  add  a  zero  to  get  one  com- 
plete group  not  all  zeros.  Take  the  square  root  of  the  largest 
square  contained  in  the  first  group  which  is  not  all  zeros  and 
prefix  to  it  as  many  zeros  as  there  are  groups  preceding  this  one, 
thus  locating  the  decimal  point  in  the  root. 


182  SQUARE   ROOTS   AND   RADICALS 

Thus,  in  the  root  of  .03  42  the  first  term  is  .1,  since  1  is  the  largest 
square  in  3.  Similarly  in  the  square  root  of  .00  70  the  first  term  is 
.08,  since  64  is  the  largest  square  in  70  and  there  is  one  group  preced- 
ing this  one. 

Illustrative  Example.     Find  the  square  root  of  .06784. 

Sohition. 

Sqttaee  Boot  Square 

a  +  b     +  c 


)6  +  .0004  =  .2604 

.06783 

a2  = 

.04 

2ab  =  .024 

.02783 

62  =  .0036 

.0276 

.0276 

2(a  +  6)6-  =  .000208 

.00023 

c2  =  .00000016 

.00020816 

.00020816 

.00002194 

Explanation.  According  to  the  rule,  .2  is  the  first  term  of  the 
root  since  4  is  the  largest  square  in  6  and  there  is  no  group  preceding 
.06.  The  process  is  the  same  as  in  the  case  of  an  integral  square,  but 
special  care  is  now  needed  in  handling  the  decimal  points,  which  is 
done  exactly  as  in  operations  upon  decimals  in  arithmetic. 

For  instance,  in  finding  the  third  term  in  this  example,  we  divide 
.00023  by  2(.26)  =  .52  and  the  quotient  lies  between  .0004  and  .0005. 
Hence  c  =  .0004. 

176.  Evidently  the  process  in  this  example  may  be  carried 
on  indefinitely.  .2604  is  an  approximation  to  the  square  root 
of  .06783.  In  fact,  the  square  of  .2604  difPers  from  .06783  by 
only  .00002184.  .260  is  the  nearest  approximation  using  three 
decimal  places.  If  the  fourth  figure  were  5,  or  any  digit 
greater  than  5,  then  .261  would  be  the  nearest  approximation 
using  three  decimal  places.  Hence,  four  places  must  be  found 
in  order  to  be  sure  of  the  nearest  approximation  to  three 
places. 


SQUARE   ROOTS   OF  FRACTIONS  183 

EXERCISES 

Find  the  square  roots  of  the  following,  correct  to  two  deci- 
mal places  : 

1.  387.  7.  2.  13.  .02. 

2.  5276.  8.  3.  14.  .003. 

3.  2.92.  9.  5.  15.  .5. 

4.  27.29.  10.  7.  16.  .005. 

5.  51.  11.  8.  17.  .307. 

6.  3.824.  12.  11.  18.  200.002. 

SQUARE  ROOTS  OP  FRACTIONS 

177.    A  fraction  is  squared  by  squaring  its  numerator  and  its 

denominator   separately,  since  -  x  -  =  -2*     Hence,  to  extract 

b      b      b 

the  square  root  of  a  fraction,  we  find  the  square  root  of  its 
numerator  and  its  denominator  separately. 

E.g.  v^  =  i  since  f  x  f  =  if- 

However,  in  approximating  the  square  root  of  a  fraction 
whose  denominator  is  not  a  perfect  square,  the  fraction  may 
be  reduced  to  a  decimal  before  the  root  is  approximated. 

E.g.  Vf  =  V.666  .  •  .  =  .8165  •  •  •  • 

It  is  also  sometimes  convenient  to  make  the  denominator  of 
a  fraction  a  perfect  square  before  approximating  the  root. 

E.g.     Vf  =  V|  =  Vf~6  =  Vl-y/Q  =  lV^  =  ?:^  =  .8165  .... 

o 

Note  that  Principle  XVIII  is  used  in  taking  the  step, 

VfTB  =  VJ  .  V6. 

It  is  clear  that  any  fraction  can  be  changed  into  an  equal  frac- 
tion whose  denominator  is  a  perfect  square  by  multiplying  nuwr 
erator  and  denominator  by  the  proper  number. 

E.g.  i  =  |,    |  =  H.    i  =  il,etc. 


184  SQUARE  ROOTS  AND   RADICALS 

178.  Example.  Find  the  square  root  of  |  by  finding  the 
roots  of  2  and  3  separately  and  then  dividing  to  reduce  to  a 
decimal. 

Compare  this  with  the  two  methods  given  above  and  show 
in  what  respect  they  are  simpler. 

179.  If  it  is  required  to  approximate  the  value  of  — -  ,  the 
simplest  method  is  as  follows  :  ^ 

_1_^_VL^V5.     Similarly,  A  =  J^.3V5. 
V5      V5V5       5  V5      V5V5        5 

Note  that  V5  VS  =  5  by  Principle  XVIII  read  in  reverse 
order. 

Thus,  Vo  V5  =  V25  =  5. 

EXERCISES 

State  three  rules  for  finding  the  square  root  of  a  fraction. 

Find  approximately  correct  to  two  decimal  places  the  follow- 
ing square  roots,  using  the  method  which  involves  the  least 
computation. 


1.  V|. 

7. 

v^. 

13. 

vi. 

2.    V|. 

8. 

VI?. 

14. 

Vf. 

3.    V|. 

9. 

V|. 

15. 

v^. 

4.    Vfl. 

10. 

Vf. 

16. 

V|. 

5.    V|. 

11. 

V-2:^. 

17. 

1 

6.    V|. 

12. 

v^. 

V6 

18. 
19. 
20. 


V13 
3 

V7 
7 


Vi7 

21.    Vf. 


SIMPLIFYING  RADICALS 
180.   Principle  XVIII   may  also  be  used  to  advantage  in 
approximating  the  square  roots  of  certain  integral  numbers. 

E.g.     Suppose  \/2  has  been  computed,  and  Vs  is  desired.    It  is 
unnecessary  to  compute  the  Vs  directly,  for  by  XVIII, 

V8  =  V4^  =  Vi  .  V2  =  2  V2. 


SIMPLIFYING   RADICALS  185 

This  sort  of  simplification  is  possible  whenever  the  number 
under  the  radical  sign  can  be  resolved  into  two  factors,  one  of 
which  is  a  perfect  square. 

E.g.    Suppose  \/5  to  have  been  computed,  then 

Vl25  =  V25T5  =  \/25  .  V5  =  5  Vs. 
In  like  manner,  y/a^b^  may  be  written 

Va*62 .  ah  =  Va*E^  •  ^/ab  =  a^b  Vab. 
181.   Definition.     An  expression  in  one  of  the  forms  V^ 
—p,  \/-,  is  said  to  be  simplified  when  it  is  reduced  so  that  no 

radical  occurs  in  a  denominator  and  when  the  number  under  the 
radical  sign  is  in  the  integral  form  and  contains  no  factor 
which  is  a  perfect  square. 

Such  radical  expressions  may  always  be  simplified  by 
Principle  XVIII. 

E.g.    .  \/i25  =  V25T5  =  5  V5 

Vo^  =  Va^ft-*  •  ah  =  a%  Vah 


V3      V3  •  V3       3 

EXERCISES 

Given  V2  =  1.4142,  V3  =  1.7321,  V5  =  2.2361,  compute  the 
following,  correct  to  three  places  of  decimals,  without  further 
extraction  of  roots : 


1. 

V80. 

6. 

V2.3. 

11. 

V27+V|. 

2. 

Vi. 

7. 

V72. 

12. 

V45  4-Vi. 

3. 

Vf 

8. 

V98. 

13. 

V50-VI+V8. 

4. 

V48. 

9. 

V363. 

14. 

V48+V75-V3. 

5. 

V75. 

10. 

V125. 

15. 

V32+V72-VI8. 

186  SQUARE   ROOTS   AND  RADICALS 

Simplify  the  following : 

16.  V32^.  19.    V45"^yP.         22.    V500  x'a^b. 

17.  V81  afb\  20.    V63  bc'd'.  23.    V3a^-f  6a;?/-f  S^/'. 


18.    V50a*6V.  21.    V900a6V.         24.    V8a;2-12/. 

25.    V32  0^2  _  64  a6  +  32  b\  26.    Vl25  ic^  -|-  250  xy  + 125  /. 


♦ 


27.  Find  approximately  to  two  decimal  places  the  sides  of 
a  square  whose  area  is  120. 

28.  Approximate  to  two  decimals  the  side  of  a  square  having 
an  area  equal  to  that  of  a  rectangle  whose  sides  are  15  and  20. 

29.  How  many  rods  of  fence  are  required  to  fence  a  square 
piece  of  land  containing  50  acres,  each  acre  containing  160 
square  rods  ? 

30.  A  square  checkerboard  has  an  area  of  324  square  inches. 
What  are  its  dimensions  ? 

In  adding  or  subtracting  expressions  containing  radicals  it  is 

always  best  to  first  reduce  each  radical  expression  to  its  sim- 
plest form,  since  this  often  gives  opportunity  to  combine  terms 
which  are  similar  with  respect  to  some  radical  expression. 

Ex.  1.  V32  4-  V72  -  V18  =  4  V2  +  6  V2  -  3  V2  =  7  V2  by 
Principles  XVIII  and  I. 

Ex.2.     V|+Vl2-V|  =  iV34-2V3-|V3 

=  a  +  2-i)V3  =  l|V3. 

EXERCISES 

Simplify  each  of  the  following  as  far  as  possible  without 
approximating  roots. 

1.  V27  H- 2  V48  -  3  V7"J.     3.   3  V432  -  4  V3  +  Vl47. 

2.  V20  +  Vi25-Vr80.       4.   3  V2450  -  25  V2  +  4  V13122. 


^  EQUATIONS   SOLVED   BY   SQUARE   ROOTS  187 

_^r 

6.  V4  3?y  4-  V25  xif  —  x  ^xy. 

7.  ->Ja:i?  —  hx^-\-  V4  ai^s^  —  4  67^s^. 

8.  4V|-fVS-^V27. 

.9.  2V|  +  V60  +  V|.  10.   5V3-2V48  +  7Vi08. 

11.  -^oT^^^^  -  Vab^ -W--yJ{a^b) {a" -  b% 

12.  Va  +  3V2a-2V3^  +  V4a-V8a  +  Vi2a. 

13.  Var*  +  2 ic^y ^ 2^2 _ ^aj3 ^2x'y-^xf  _ V4^. 

14.  Vr-sH-Vl6r-16s  +  Vri^-s«2_V9(r  — s). 

15.  V(m  —  w)^a  +  V(w  +  ?i)^a  —  y/aiii^  +  Va  (1  —  m)^  —  Va. 

16.  V32^*  -f  Vl62  a^y  -  V512  ar^/  H-  V1250  a^«/\    . 

EQUATIONS   SOLVED  BY  SQUARE  ROOTS 
182.    Since  2^=4  and  also  (-2)2  =  4,  it  follows  that  the 

equation  a^=4  has  two  roots,  namely  a; =2  and  a;=  —  2.    These 

are  usually  written  x=  ±2. 

This  solution  is  obtained  by  taking  the  square  root  of  both 

sides,  which  is  equivalent  to  dividing  both  sides  by  the  same 

number. 

This  operation  may  now  be  added  to  those  enumerated  in 

Principle  VI  for  the  solution  of  equations. 

EXERCISES 

Find  all  roots  of  the  following  equations : 

1.  x'  =  9.  6.    a:2==G4a?AV.  11.  a^  =  98. 

2.  ar^  =  25.  7.    x"  =  36  7^8*.  12.  x^^gO. 

3.  a^=16a\  8.   a^^Sls^r^.  13.  a;2  =  175. 

4.  ar^=49  62.  9.   3^  =  e25a*b\  14.  ar^  =  49a^ 

5.  x^=  81  a'^b^         10.   aj2  =  72.  15.  a^  =  36a^b\ 

16.  a^==2o(a-by.  18.    a:^  ^  200(a  +  f^)^. 

17.  x'  =  50{a  +  b).  19.    a.'2=1250(a-6)V. 


188  SQUARE  ROOTS  AND  RADICALS 

-DRILL  EXERCISES 

Using  the  formulas  for  (a  ±  by,  obtain  the  squares  of  the 
following : 

1.  (a  +  6)  +  (c— d).         4.   7a;— (4r  — s). 

2.  (a  +  3)-(6  +  c).  5.    (m2-3)-2(7^i3  +  n). 

3.  (3a-26)  +  5.  6.   3(2  +  2/)- 2(3  + a^). 

Factor  the  following : 

7.  (2-xf-2{2-x){x-l)+{x-lf. 

8.  (2+2/)2+2(2+2/)(l  +  a;)  +  (l  +  a;)l 

9.  (3a-2?>)2-10(3a-26)+25. 

10.  (6  a  -  hf '+  (2  a  +  1)^  _  2(6  a  -  6)(2  a  + 1). 

11.  25(a  +  6)2  +  50(a  +  6)(a  -  6)  +  25(a  -  h)\ 

12.  a;2^i2a:(a  +  6  +  c)+36(a  +  6  +  c)2. 

13.  49(m  -  3)4  +  36(m  + 1)^  -  84 (m  -  3)2(m  +  1)^ 

14.  16(ic  -  2/)2  -  16(a;  -y){x^y)+  4(x  +  2/)l 

15.  -  30(a  +  6)(a  -  6)^  +  2o(a  -  6)^  +  9(a  +  hf. 

Using  the  formula  for  (a  +  &)(a  —  6),  write  out  the  following 
products : 

16.  [a  +  6  +  (c-d)][a  +  6-(c-(«)]. 

17.  lx  +  y+{u-\-v)'][_x  +  y-{u  +  v)'\. 

18.  [4a;-(a-26)][4a;  +  (a-26)]. 

19.  la  +  2h-(x-f)-]la  +  2h  +  {x-y')-\, 

20.  (11  Wx  -Sba^  (11  6^0;  +  3  6ic^. 

Factor  the  following : 

21.  a^-h4:ab-\-4:b^-(x^-2xy  +  y^. 

22.  (3a;-2)2-(4^2_^9/-12a;2/). 

23.  a^-{-4.xy-\-4:y^-(a^  +  2ab-hb^. 

24.  '16  xy  _  (4  a;2  +  9  2/2  + 12  a;?/)- 
26.    (a  +  6)2-(4a2  +  9  62-12  6c). 


APPLICATIONS  OF  SQUARE  ROOT  189 

APPLICATIONS  OP  SQUARE  ROOT 

183.  Some  of  the  most  interesting  and  useful  applications 
of  the  square  root  process  are  concerned  with  the  sides  and 
areas  of  triangles. 

The  fact  that  the  sum  of  the  squares  on  the  two  sides  of  a 
right  triangle  equals  the  square  on  the  hypotenuse  was  used 
in  Chapter  X.     (Pythagorean  Proposition,  page  172.) 

If  a  and  h  are  the  lengths  of  the  sides,  and  c  the  length  of 
the  hypotenuse,  all  measured  in  the  same  unit,  this  proposi- 
tion says:  c'=d?-^JP,  (1) 

Hence,  by /S,  a?  =  (?-h\  (2) 

and  h^  =  c'-a\  (3) 

Taking  the   square   root  of  both   sides  in  each  of  these 

equations,                           c  =  V^M^«.  (4) 

a  =  V?^=^.  (5) 

h  =  V^^^\  (6) 

The  negative  square  root  is  omitted  here,  as  a  negative 
length  cannot  apply  to  the  side  of  a  triangle.  By  these  formu- 
las, if  any  two  sides  of  a  right  triangle  are  given,  the  other 
may  be  found. 

E.g.  if  a  =  4,  &  =  3,  then,  by  (4), 

c  =  VPT32  =  V16T9  =  V25  =  5. 

If  c  =  5,  6  =  3,  then,  by  (5),       ^ 

a  =  V52  -  3-2  =  V25-9  =  Vie  =  4. 

If  c  =  5,  a  =  4,  then,  by  (6), 

I  =  V52  -  42  =  y/2d  -  IG  =Vd  =  3. 

Illustrative  Problem.  If  the  two  sides  of  a  right  triangle  are 
8  and  12,  find  the  hypotenuse  correct  to  two  decimal  places. 


Solution.     We  have       c  =  Va^  +  b^  =  \/64  +  144  =  V208, 

V2O8  =  Vi6Ti3  =  VI6  .  Vl3  =  4a/I3  =  4(3.605)  =  14.420. 


190 


SQUARE  ROOTS  AND   RADICALS 


PROBLEMS 

In  solving  the  following  problems,  simplify  each  expression 
under  the  radical  sign  before  extracting  the  root.  Find  all 
results  correct  to  two  decimal  places. 

.     1.   The  sides  about  the  right  angle  of  a  right  triangle  are 
each  15  inches.     Find  the  hypotenuse. 

2.  The  hypotenuse  of  a  right  triangle  is  9  inches  and  one 
of  the  sides  is  6  inches.     Find  the  other  side. 

3.  The  hypotenuse  of  a  right  triangle  is  25  feet  and  one 
of  the  sides  is  15  feet.     Find  the  other  side. 

4.  The  hypotenuse  of  a  right  triangle  is  7  rods  and  one  of 
the  sides  is  5  rods.     Find  the  other  side. 

5.  The  hypotenuse  of  a  right  triangle  is  12  inches  and  the 
two  sides  are  equal.     Find  their  length. 

Let  s  be  the  length  of  one  of  the  equal  sides. 
Then  ^-\-fr'=iU. 

2.s2  =  144. 
_^=72._ 
s  =  V72  =  6  V2  =  6  X  1.414  =  8.484. 

6.  The  hypotenuse  of  a  right  triangle  is  30  feet  and  the 
sides  are  equal.     Find  their  length. 

7.  The  hypotenuse  of  a  right  triangle  is  h  and  the  sides 
are  equal.  Find  their  length.  Solve  Exs.  5  and  6  by  means 
of  the  formula  here  obtained. 

8.  The  diagonal  of  a  square  is  8  feet.     Find  its  area. 

9.  The  diagonal  of  a  square  is  d.  Find  an  expression  in 
terms  of  d  representing  its  area. 

10.  The  side  of  an  equilateral  triangle 
is  6  inches.     Find  the  altitude. 

A  line  drawn  from  a  vertex  of  an  eqviilateral 
triangle  perpendicular  to  the  base  meets  the 
base  at  its  middle  point.  Hence  this  problem 
becomes :  the  hypotenuse  of  a  right  triangle  is 
6  and  one  side  is  3.     Find  the  remaining  side. 


APPLICATIONS   OF  SQUARE   ROOT  191 

11.  The   side   of  an  equilateral   triangle   is  10.     Find  the 
altitude. 

12.  The   side   of  an   equilateral  triangle   is   s.     Find  the 
altitude. 

This  is  equivalent  to  finding  a  side  of  a  right  triangle  whose  hypote- 
nuse is  s,  the  other  side  being  -  •     Let  h  equal  altitude. 


Then  h=yls^-(ty  =  ^^l 

'4  2 

This  formula  gives  the  altitude  of  any  equilateral  triangle  in  terms 
of  the  side.     By  means  of  this  formula  solve  Exs.  11  and  12. 

13.  Find  the  altitude  of  an  equilateral  triangle  whose  side 
is  4^.     Substitute  in  the  formula  under  Ex.  12. 

14.  Find  the  area  of  an  equilateral  triangle  whose  side  is  5. 

Since  the  area  of  a  triangle  is  ^  the  product  of  the  base  and  alti- 
tude, we  first  find  the  altitude  by  means  of  the  formula  under  Ex.  12, 
and  then  multiply  by  i  the  base. 

15.  Find  the  area  of  the  equilateral  triangle  whose  side  is  s. 

Show  the  result  to  be  —  V3. 
4 

16.  If  the  area  of  an  equilateral  triangle  is  16  square 
inches,  find  the  length  of  the  side. 

Let  s  equal  the  length  of  the  side.     Then  by  the  formula  derived 

in  Ex.  15,  we  have  16  =  -  V3. 
4 

.  Hence  (§§  177-180),  s^  =  ^  =  M  V3  =  21.33  x  1.732. 
•     \/3      3 

17.  The  area  of  an  equilateral  triangle  is  50  square  inches. 
Find  its  side  and  altitude. 


192  SQUARE  ROOTS  AND  RADICALS 

18.  The  area  of  an  equilaterial  triangle  is  a  square  inches. 
Find  the  side. 

Solve  the  equation  a  =  -~  V'S  for  s,  and  simplify  the  expression, 

finding    .^  =  ±?,and.  =  V^  =  |V?IVS. 

19.  The  area  of  an  equilateral  triangle  is  240  square  inches. 
Find  its  side.      (Substitute  in  the  formula  obtained  in  Ex.  18.) 

20.  Find  tl^e  area  of  a  regular  hexagon  whose 
side  is  7. 

A  regular  hexagon  is  composed  of  6  equal  equi- 
lateral triangles,  whose  sides  are  each  equal  to  the 
side  of  the  hexagon  (see  figure).  Hence  this  prob- 
lem may  be  solved  by  finding  the  area  of  an  equi- 
lateral triangle  whose  side  is  7,  and  multiplying  the  result  by  6. 

21.  Find  the  area  of  a  regular  hexagon  whose  side  is  s. 
Solve  Ex.  20  by  substituting  in  the  formula  obtained  here. 

22.  The  area  of  a  regular  hexagon  is  108  square  inches. 
Find  its  side. 

If  the  area  of  the  hexagon  is  108  square  inches,  the  area  of  one  of 
the  equilateral  triangles  is  18  square  inches.  Hence  this  problem  can 
be  solved  like  Ex.  18. 

23.  The  area  of  a  regular  hexagon  is  a  square  inches.  Find 
its  side.  Solve  Ex.  22  by  substituting  in  the  formula  obtained 
here.  ^^s.    s  =  iV2aV3. 

24.  Find  the  radius  of  a  circle  whose  area  is  9  square  inches. 

The  area  of  a  circle  is  found  by  squaring  the  radius  and  multiply- 
ing by  3.1416.  The  number  3.1416  is  approximately  the  quotient 
obtained  by  dividing  the  length  of  the  circumference  by  the  diameter 
of  the  circje.  This  quotient  is  represented  by  the  Greek  letter  tt 
(pronounced  pi).  In  this  chapter  we  use  3f  as  an  approximation  to  tt. 
This  differs  from  the  real  value  of  tt  by  less  than  .0013,  and  hence  is 
accurate  enough  for  most  purposes.  If  -a  represents  the  area  of  a 
circle,  the  above  rule  may  be  written 


FURTHER  OPERATIONS   ON  RADICALS  193 


Hence  if  a=  9,  r^  =^  =  |-  =  ^  =  2.863, 


and  r  =  V2.863. 

25.  Find  the  radius  of  a  circle  whose  area  is  68  square  feet. 

26.  Find  the  radius  of  a  circle  whose  area  is  a  square  feet. 
We  have  a  —  irr^,  or  r^  =  - . 

Hence  r  =  V^  =  V^=iV^. 

In  problems  stated  in  terms  of  letters,  the  results,  of  course,  cannot 
be  reduced  to  a  decimal.  In  such  formulas  it  is  best  not  to  replace 
the  letter  tt  by  any  of  its  approximations. 

FURTHER  OPERATIONS  ON  RADICALS 

184.  The  radical  sign  is  used  to  indicate  other  roots  than 
square  roots  by  means  of  an  index  figure. 

Thus,  the  cube  root  of  8,  or  one  of  its  three  equal  factors,  is  written 
■v^8  =  2.  The  fourth  root  of  16,  or  one  of  its  four  equal  factors,  is 
written  v^l6  =  2. 

185.  Definitions.  Any  expression  which  contains  an  indi- 
cated root  is  caljed  a  radical  expression. 

Integers  and  fractions  of  the  form  —  where  m  and  n  are  in- 
tegers  are  called  rational  numbers. 

E.g.  2  +  \/5  is  a  radical  expression.  5,  §,  -^^j^^  are  rational  num- 
bers. 

186.  An  expression  which  consists  of  a  rational  number 
under  a  radical  sign,  or  one  which  can  be  reduced  to  this 
form,  is  called  a  surd,  provided  the  whole  expression  is  not  re- 
ducible to  a  rational  number. 

E.g.  V2  is  a  surd  since  it  cannot  be  reduced  to  a  rational  num- 
ber. .  V4»  \/3  are  surds  for  the  same  reason.  VQ  is  not  a  surd 
since  \/9  =  3.  V2  -f  y/2  is  not  a  surd  since  2  +  "\/2  is  not  a  rational 
number. 


194  SQUAKE  ROOTS  AND  RADICALS 

187.  Surd  expressions  containing  indicated  square  roots  only, 

and  in  which  no  number  is  under  more  than  one  radical  sign, 

are  called  quadratic  surds. 

_         —              _  3 

E.g.      V2  +  V3,   3  +  VS,    — = t=,   are   quadratic   surds,  while 

v7  —  V  5 

V  V2  and  V4  are  not  quadratic  surds. 

The  following  operations  upon  radicals  deal  only  with  quad- 
ratic surds. 

MULTI -'LIGATION  OF  QUADRATIC  SURDS 

188.  By  Principle  XVIII  V^  =  Va  •  V6. 

This  equation  read  in  the  reverse  order  gives  a  rule  for  mul- 
tiplying quadratic  surds. 

E.g.      V2  .  V8  =  v^Ts  =  Vl6  =  4. 

V3  .  VI5  =  V45  =  \/9~^  =  3  V5. 

EXERCISES 

Make  a  rule  for  multiplying  quadratic  surds. 
Perform  the  following  indicated  operations  and  reduce  each 
result  to  the  simplest  form. 

1.  V5.V7.  4.    V^-V^.  ,7.    V7.Vf 

2.  V6.V12.  5.    V^-V^l  8.    Vf-VI. 

3.  V18.V8.  6.    V^^-V?^.  9.    Vf-Vf 

10.  (V2  +  V3)2. 

Solution.     ( V2  +  \/3)(V2  +  V3)  =  ( V2)2+2  V2\/3  +  ( V3)2  = 
4  +  2  V6  +  3  =  7  V  2  V6 .     • 

11.  (V2-V3)2.  16.  (V^4-V6)(V^-V6). 

12.  (H-V2)2.  17^  V2(2V3  +  3V8-5V6). 

13.  (2-V3)2.  18.  Va6c(VaH-V6H-Vc). 

14.  (H-V2)(1-a/2).  19.  (3V5-2V3)(3V5+2V3). 

15.  (V2  +  V5)(V2-V5).  20.  (l-hV2)(l-V3+V5). 


DIVISION  OF   QUADRATIC   SURDS  196 

DIVISION  OF   QUADRATIC   SURDS 

189.  When  a  divisor  is  a  quadratic  surd,  it  is  convenient  to 
indicate  the  division  in  the  form  of  a  fraction  and  then  to 
reduce  the  denominators  to  the  rational  form  as  in  the  follow- 
ing example : 

V5      _        VE(V5  +  V2)       ^  5  +  VIo_  s  +  VTo 
V'5-V2      (V5-V2)(V5  +  v/2)        5-2    ~        3 

If  it  is  desired  to  compute  the  approximate  value  of 
V5"^(V5— V2),  it  obviously  requires  less  numerical  work 

to  use  the  form      "*"    —  rather  than    — — ;   since   the 

3  V5-V2 

latter  involves  the  extraction  of  two  square  roots  and  a  long 
division,  while  the  former  requires  the  extraction  of  only  one 
root  and  a  short  division. 

190.  The  process  indicated  in  the  above  example  is  called 
rationalizing  the  denominator,  and  the  factor  by  which  the 
terms  of  the  fraction  are  multiplied  is  called  the  rationalizing 
factor. 

If  the  denominator  is  of  the  form  Vx  or  aVx,  then  Vi  is  the 
rationalizing  factor,  since  Vx  '  \/x  =  x. 

If  the  denominator  is  of  the  form  V3:  +  Vy,  then  y/x  —  Vy  is  the 
rationalizing  factor,  since  (  Vx  +  Vy)(Vx  —  y/y)  =  x  -  y. 

Give  a  rationalizing  factor  of  each  of  the  following : 

.     (1)  V3^,         (2)  V2^,         (3)  V8'6V,         (4)  Va  +  V^ 

(5)  Va-V6,         (6)  Vk^  +  V2T^         (7)  V7-V27. 

191.  To   rationalize  the  denominator   of    any   fraction,    it   is 

first  necessary  to  find  an  expression  which  multiplied  by  the 
denominator  of  the  fraction  gives  a  rational  product,  and  then 
to  multiply  both  terms  of  the  fraction  by  this  expression. 


196  SQUARE  ROOTS  AND  RADICALS 

EXERCISES 

Rationalize  the  denominators  of  each  of  the  following : 

1    ^.  8.   3V6  +  9V2  ^g  Va 

'   Va;  _  3V2  "  Va-V6 


Va;  ■   ^/9_1*  16. 


^  Va  +  V6 


3V2 

Q 

1 

V2-1 

1  0 

1 

V5-V3 

1  1 

1 

V3+V5 

1  9 

2 

V7-V3 

13. 

2 
Va-1 

1/1 

1 

V6 


^^  +  1  •   V5-V3  17^   Va-V6 

4.  V^  — ^.  1  Va  +  V6 

^/^tV  '''^''  18     V^  +  V^ 

5.  ^^^-  12.    -^ V5WF 

Va;2/ 

1 


6    V2  +  V3  2  19.  -      _ 

V2  'Va-1                         aVa?+W2/ 

7.    V8--^^  ,4        _1      _.  20.          g+^     _■ 

V3  Va  +  V6                     a^/x  —  b^y 


EQUATIONS  INVOLVING  RADICALS 
192.    Illustrative  Example.     Solve  for  x  the  equation ; 


Vx-5+Vx  +  l  =  3  (1) 

By5|VJTl  V^35  =  3-\/^Tl  (2) 

Squaring  both  sides  a;— 5  =  9  — 6Va;  +  l  +  a:  +  l  (3) 

SimpHfying  2  Va:  + 1  =  5  (4) 

Squaring  both  sides  4(a;  +  1)  =  25'  (5) 

Hence  x  =  5J  (6)* 

Check.     Substitute  x  =  5^  in  (1) 

If  two  radicals  are  involved  or  one  radical  and  rational 
terms,  it  is  best  to  get  a  radical  alone  on  one  side  of  the^  equa- 
tion before  squaring  as  in  (2)  and  (4).  Note  that  squaring 
both  members  of  an  equation  is  equivalent  to  multiplying  both 
sides  by  the  same  number. 


EQUATIONS   INVOLVING  RADICALS  197 

EXERCISES 

Solve  the  following  equations  and  check  each  result. 


1.  Va;  — 5  =  3.  7.    V4ar*  — 7H-2a;  =  7. 


2.  aj-2=Va^-4.  8.    Va;-f  1  - Va;- 7  =  2. 


3.  Va;  +  6  =  3Va;-2.  9.    V2/  +  4  + V?/- 1  =  5. 

4.  Va^ -2a;  +  8  =  a;-4.  10.    V^  +  2  =  Va; -f  16. 


5.  Va^  -f  7  a;  —  4  =  Va.'2+  8  a;  —  5.   11.   5  —  Va;  =  Va;  +  5. 


6.  V^+5  +  «  =  5.  12.    Va;4-7  =  l+Va;H-2. 

j3    3(-v^-2)^V^  +  l  14.    V^^^^  V^+1^ 

V^— 1        V^4-2  Vx  — 4      Va;  — 3 

15.        i       +-J: =  ^L_.  le     2-V^^3+V^ 

Va;H-l      Va;-1     a;-l  *    ^^^^     3-V^' 

Suggestions.     In  Ex.  13  clear  of  fractions  first;  in  14  square  both 
members  and  then  clear  of  fractions ;  in  15  rationalize  denominators. 

REyiEW  QUESTIONS 

1.  State  Principle  XVIII.     Show  by  use  of  this  principle 
how  to  find  V28  having  given  Vf  =  2.696. 

2.  Show  how  the  value  of  the  following  may  be  approxi- 
mated by  finding  only  one  square  root. 

5V20  +  2V45  -  3V80  +2Vi. 

3.  Write  the  square  of  a  -|- 6  +  c + d  in  such  a  form  as  to  derive 
from  it  the  rule  for  finding  the  square  root  of  a  polynomial. 

4.  When  is  a  radical  expression  said  to  be  simplified?    How 
is  Principle  XVIII  used  for  this  purpose  ? 

5.  Give  examples  of  rational  expressions,  of  surds,  of  quad- 
ratic surds.     When  is  an  expression  said  to  be  rationalized  ? 

6.  What  principle  is  used  in  multiplying  two  quadratic 
surds  ?     Find  the  product  (1  +  V2)(2+  3  V2). 

7.  How   is    division    by   a    quadratic    surd    carried   out? 
Simplify  2 V3  -- (V2  -  V3) ;  also  3 V6  -^( V5  -  V2). 


198  SQUARE   ROOTS  AND   RADICALS 

8.  In  solving  an  equation  containing  radicals,  how  are  they 
removed  (1)  when  only  one  radical  expression  is  involved ; 
(2)  when  two  such  expressions  are  involved  ? 

9.  State  Principle  XVIII  in  symbols  and  thus  complete 
your  list.     Put  them  all  on  a  single  sheet  for  handy  reference. 

DRILL  EXERCISES 

1.  Divide  a^ - 12  a2+ 27  a +  40  by  a -5. 

2.  Dividear'-5aj*2/-hllaj3/-14a^z/3+9a;y-2/byi«2-3ri/ 

3.  Diwide  x^  +  x^y^ -^y*hy  a^  —  xy  +  y^. 

4.  Divide  a^-^5a^ -2  a-24:  by  a^-f  7a.4-12. 

5.  Divide  a'- 5  a^b -\-10  a^b' -10  a'b^  + 5  ah'- b'  by  a' - 
2ab-\-b\ 

6.  Divide  ar'  —  5 a^y  —  5 x^i^  +  2/*  by  x^—^xy  +  y^. 
Factor : 

7.  a^+S.  13.   1  +  64^3,  19    i_|_i25a;«. 

8.  21a?^b\  14.    w^-\-21a\  20.    27a;«-l. 

9.  63_27.  15.   w^-^a\  21.   l-^^y\ 

10.  8a«-&l  16.    27 a^- 8 63.  22.    l  +  ^^y\ 

11.  l  +  64a;^  17.    :i^-\-y\  23.    %o^^27y\ 

12.  a^-b\  18,   125 a'' +  6^  24.    ^:x?~21y\ 

25.  c*-31c2  +  220.  27.    26  +  39  n-22m-33mw. 

26.  ac  +  d'a  -  b'c  -  b'd\  28.   12  ar*  + 11  a;  -  56. 

29.  a2  +  4a6  +  462_(a2-4a6  +  4&2). 

30.  (3aj-l)2-(a;2  +  42/'-4a^). 

31.  (a;  +  32/y  +  (a;-22/)2  +  2(a;+32/)(a:-22/). 

32.  16  (a  +  6)2-  8  (a  -  6)(a  +  6)  +  (a  -  6^. 

33.  256ic2-(49ar'  +  42/*-28a;2/2). 

34.  (2a;-a)2  +  100(a-3a;)2+20(2a;-a)(a-3a;). 

35.  -  48  (a  -  b)ia  +  6)  +  36  (a  -  bf  +16  (a  +  b)K 


CHAPTER  XII 
QUADRATIC  EQUATIONS 

193.  Equations  of  the  form  x^ -{- ax  -\- b  =  0  have  already 
been  solved  in  cases  where  the  left  members  could  be  factored 
by  inspection.  However,  in  a  case  like  a^  +  5a;-f3=0,  the 
factors  of  the  left  member  cannot  be  found  by  any  method 
thus  far  studied.  It  is  therefore  necessary  to  consider  other 
methods  for  solving  equations  of  this  type. 

194.  As  a  preliminary  step  let  us  consider  again  the  proper- 
ties of  a  trinomial  square. 

How  many  terms  of  such  a  trinomial  must  be  squares  ?  How 
is  the  remaining  term  related  to  these  ?  What  term  must  be 
added  to  a^  -f-  2  ab  in  order  to  make  this  a  trinomial  square  ? 

The  process  of  finding  this  third  term  is  called  completing 
the  square.  Since  2  ab  must  be  twice  the  product  of  the  square 
roots  of  the  squared  terms,  it  follows  that  b  of  the  missing  term 
may  be  found  by  dividing  2  ab  by  tivice  a. 

EXERCISES 

Complete  the  trinomial  square  in  each  of  the  following : 

1.  x'-\-2x.  4.   a^-\-Sx.  7.    {Sxy-\-2(Sx). 

2.  x^  +  4:X.  5.    x^  +  Sx.  8.    (2a;)2+4(2a;). 

3.  0^  +  6  a*.  6.   x^-\-5x.  9.   16  of  +  2(4.  x). 

10.  How  do  you  complete  the  square  in  a^  —  2ab?  Is  the 
rule  different  in  this  case  ? 

11.  Complete  the  square  in  each  of  the  above  exercises,  first 
replacing  the  sign  +  by  — . 

199 


200  QUADRATIC   EQUATIONS 

195.   Solution  of  a  quadratic  by  completing  the  square. 
Ex.1.    Solve  the  equation : 

x'  +  6x-\-4:  =  0.  (1) 

By  5  I  4,  a;2  +  6  a:  =  -  4.  (2) 

Adding  3^  to  both  members  to  complete  the  square, 

x^  +  Qx  +  8^  =  S-^-4:  =  5.  (3) 

Taking  square  root  of  both  sides,  x  +  3  =  ±  V5.  (4) 

Hence  a:  =  -3  +  V5  =  -3  +  2.24  =  -  .76, 

and  .  a:  =  -  3  -  V5  =  -  3  -  2.24  =  -  5.24. 

Ex  2.   Solve  the  equation : 

a^-12x  +  4:2  =  56.  (1) 

By  S,  a;2  -  12  x  =  14.  (2) 

Completing  the  square,   a;^  -  12  a:  +  36  =  14  +  36  =  50.  (3) 

Taking  square  roots,  x  -  Q  =  ±  VEO  =  ±  5  V2.  (4) 

By  ^,  x  =  Q±  7.071.  (5) 

Hence  a:  =  6  +  7.071  =  13.071, 

and  also  a:  =  6  -  7.071  =  -  1.071. 

This  process  is  called  solving  the  quadratic  equation  by 
completing  the  square. 

Make  a  rule  for  solving  a  quadratic  equation  by  this  process. 

EXERCISES 

In  solving  the  following  quadratic  equations  the  result  may 
in  each  case  be  reduced  so  that  the  number  remaining  under 
the  radical  sign  shall  be  2,  3,  or  5.  (§  180.)  Use  V2= 1.414, 
V3  =  1.732,  V5  =  2.236. 

1.  x^-4:X  =  S.  6.  a;2-12a;  =  12.  11.  S  =  x^-^4:X. 

2.  x'  =  S-6x.  7.  x^-Sx  =  -U.  12.  2S-6x  =  x'. 
8.  4a;=16-ar^.  8.  a^=2a;H-l.  13.  7+2a;  =  a;l 
4.  x^-\-6x  =  9.  9.  x^-4:X  =  16.  14.  25-a^  =  5x. 
6.  a;2-f.6a;  =  ll.  10.  o?  =  2^-\-^x.  15.  a^  +  ia;  =  2. 

16.    x''-^x  =  ^-. 


QUADRATIC   EQUATIONS  201 

196.  The  Hindu  method  of  completing  the  square.  In  case  the 
coefficient  of  x^  is  not  unity,  as  in  3  a^  +  8  a;  4-  4,  both  members 
may  be  divided  by  this  coefficient,  and  the  solution  is  then 
like  that  of  Exs.  1  and  2  above. 

However,  the  following  method  is  sometimes  desirable : 

3x2  +  8x  =  4.  (1) 
Multiplying  each  member  of  (1)  by  4  •  3  =  12, 

36x2  + 96  a:  =  48.  (2) 
Completing  the  square, 

36a.-2  +  96  a:  +  82  =  48  +  64  =  112.  (3) 

Taking  square  roots,        6 a:  +  8  =  ±  V\\2  =  ±  4 V7.  (4) 

Hence  a;  =  -f±4V7.  (5) 

The  advantage  of  this  form  of  solution  is  that  fractions  are 
avoided  until  the  last  step,  and  the  number  added  to  complete 
the  square  is  the  square  of  the  coefficient  of  x  in  the  original 
equation. 

This  is  called  the  Hindu  method  of  completing  the  square. 

Note.  Fractions  would  also  be  avoided  in  the  above  solution  if 
equation  (1)  were  multiplied  by  3  instead  of  4  •  3.  This  is  the  case 
only  when  the  coefficient  of  x  is  an  even  number. 

EXERCISES 

In  the  solution  of  the  following  equations  the  roots  which 
contain  surds  may  be  left  in  simplified  radical  form. 

1.  2a^  +  3a;  =  2.  9.  4a^  =  2a;  +  l.  17.  2x  +  ^x'^  =  ^. 

2.  3ar^  +  5a;  =  2.  10.  6.^-l  =  3ar'.  18.  4a^-l  =  3x. 

3.  3a;  =  9-2a5*.  11.  2a^  +  4a;  =  23.  19.  4.x  =  7 -2s?. 

4.  6ir  +  l=:-3ar'.  12.  3.r*-7  =  4a;.  20.  2a;  +  l  =  5a^. 

5.  2ar^  =  5a;  +  3.  13.  2x^-b  =  ^x.  21.  3 rc^  +  4 a;  =  7. 

6.  4x  =  2ic2-l.  14.  4a^  =  6.'c-l.  22.  3aj  +  9  =  2a^. 

7.  2a^-3a;  =  14.  15.  2a;  =  l-5a^.  23.  2a;-l  =  -4a;2. 

8.  3ic2  =  9H-2x.  16.  ^x-20=-2x\  24.  5a^+16a;=-2. 


202  QUADRATIC   EQUATIONS 

CHECKING  RESULTS   IN   QUADRATIC  EQUATIONS 

197.  Illustrative  Examples.  1.  Solving  a;^  — 7  a; +  12  =  0,  we 
get  x  =  4:  and  x  =  3. 

What  is  the  sum  of  these  roots?  How  does  this  compare 
with  the  coefficient  ot  x?  What  is  the  product  of  these 
roots?  How  does  this  compare  with  the  known  term  of  the 
equation  ? 

2.   Solving  x^—6x-^4:=0,  we  get  a;=3  +  V5  and  x=3  —  Vo. 

What  is  the  sum  of  these  roots  ?  How  does  this  sum  com- 
pare with  the  coefficient  of  a;  ?  What  is  the  product  of  these 
roots?  How  does  this  product  compare  with  the  known  term 
of  the  equation  ? 

Solve  the  following  equations.  In  each  case  compare  the 
product  of  the  roots  with  the  known  term  and  the  sum  of  the 
roots  with  the  coefficient  of  ic. 

1.  a;2-5aj  +  3  =  0.  4.  x''-4.x-^=0. 

2.  3.'2  +  3aj  +  2  =  0.  5.  x^^^x-3  =  0. 

3.  cc^  +  9a;-f  8  =  0.  6.  aj2-8aj='6. 

198.  These  exercises  are  illustrations  of  a  general  rule  for 
all  quadratics  written  in  the  form  jr^  -f  /7jr  +  y  =  0,  in  which 
the  coefficient  of  the  squared  term  is  + 1,  and  all  terms  are 
transposed  to  the  left  member ;  namely  : 

The  sum  of  the  roots  is  equal  to  the  coefficient  of  x  with  its  sign 
changed  and  the  product  of  the  roots  is  equal  to  the  known  term. 

This  may  be  used  to  check  the  results  obtained  in  solving  a 
quadratic.  Note  in  particular  that  before  applying  the  test  you 
must  put  the  equation  into  the  specified  form. 

Note  that  in  the  radical  form  the  product  of  the  roots  is  exactly 
the  known  teriii,  but  when  the  roots  are  approximated  in  the  decimal 
form,  then  their  product  is  only  approximately  equal  to  the  known 
term. 


SOLUTION   BY    FORMULA  208 

EXERCISES 

Solve  the  following  equations  and  check  all  results  by  means 
of  §  198. 

1.  4.x^-\-l=Sx.  13.  3x'  +  2x  =  5. 

2.  2a;2_3^^20.  14.  2  +  3a;=2ar. 

3.  2x^-S  =  -5x.  15.  Sx-{-l  =  -4:x'. 

4.  3x2  +  4a;  =  8.  16.  S-\-4.x==zSx^. 

5.  10-4.x=5x'.  17.  H)-\-4:X  =  5x'. 

6.  l-h4:x'  =  -6x.  18.  2  +  5x=3x'. 

7.  5-3a;  =  2a^.  19.  3x-{-U  =  2x'.- 

8.  74-4a;  =  2ar'.  20.  3a^-2a;  =  5. 

9.  6a^-f  12a;  =  2.  21.  2aj2  +  4ic  =  l. 

10.  6a^-12a;  =  -2.  22.    4a;2  +  3a;  =  l. 

11.  6a^  +  12aj  =  -2.  23.   2a^-4a;  =  23. 

12.  6x2_i2a;  =  2.  24.    2a:2_3a;  =  i. 

SOLUTION   OP   THE    QUADRATIC    BY   FORMULA 

199.   Solve  the  equation 

ax'-{-bx^c  =  0.  (1) 

By  S,  M,  4  n^x^  +  4  aftx  =  -  4  ac.  (2) 

Completing  the  square,  4  a^x^  -f  4  o6a:  +  ft^  _  52  _  4  qc.  (3) 

Taking  square  roots,  2ax  +  b  =  ±  Vb^  —  4  ac.  (4) 

hyS,D,  ■  ,^-b±VI,^-*ac_  ^.^ 

Calling  the  two  values  of  x  in  the  result  x^  and  ccg  we  have, 


Verify  these  results  by  means  of  §  198. 

Any  quadratic  equation  may  be  reduced  to  the  form  of  (1) 
by  simplifying  and  collecting  the  coefficients  of  x^  and  x. 
Hence  any  quadratic  equation  may  be  solved  by  substituting 
in  the  formulas  just  obtained. 


204  QUADRATIC  EQUATIONS 

Example.    Solve     2x^ -4:X  -\-l=0. 
In  this  case  a  =  2,  &  =  —  4,  c  =  1. 


Hence 
From  which 


•2.2 

2  +  \/2      ^       2  -  V2 


2      '  2 

Check  the  results  by  §  198. 

200.  A  quadratic  equation  may  be  proposed  for  solution 
which  has  no  roots  expressible  in  terms  of  the  numbers  of 
arithmetic  or  algebra  thus  far  studied. 

Example.     Solve  a;2  +  4  a;  =  -  8.  (1) 

By^,  a;2+4x  +  4=:-4.  (2) 

Taking  square  roots,  x  +  2  =  ±  V— 4.  (3) 

V— 4  is  unknown  to  us  as  a  number  symbol,  since  there  is  no 
number  thus  far  considered  whose  square  equals  —4.  (See  Principle 
IX.)  Such  symbols  are  defined  and  used  in  the  Advanced  Course, 
and  are  called  imaginary  numbers.  Any  quadratic  equation  which 
gives  rise  to  such  a  solution  is  to  be  interpreted  as  stating  some  con- 
dition not  satisfied  by  any  number  so  far  studied. 

EXERCISES 

Leave  the  surds  or  imaginaries  in  simplified  radical  form. 

1.  7-Sx  =  5a^.  7.  0^2- 8  4- 3  a;  =  -15  aj. 

2.  51a;-33  =  3a;2^  ,      8.  11  x^  -  49a;  +  57  =  0. 

3.  Ux-{-S-x'  =  52-Sa^.  9.  3  a;^.^  13  _iea;  =  5. 

4.  12-51a;  =  36  +  6a;2^  10.  37- 4a;2- 12a;  =  79-5flj2 

5.  5a;  +  a;2  4-8  =  0.  11.  10  ar^  +  41 +  7a;  =  44. 

6.  5aP-Slx  =  -6.  12.  45-f 3a;2-85-2a;  =  0. 

MISCELLANEOUS  QUADRATICS 
Solve  as  many  as  possible  of  the  following  equations  by  fac- 
toring.    When  this  is  not  convenient,  use  the  formula  of  §  199, 
or  complete  the  square  independently  in  each  case.     It  is  im- 
portant to  check  the  results  in  the  radical  form  as  in  §  198. 


MISCELLANEOUS  QUADRATICS  206 

1.  ic2  4.11a;  =  210.  21.  2  a^  +  3  x-3  =  12  x-{-2. 

2.  5x^-3x  =  4:.  22.  3a^--7x  =  10. 

3.  7ic  +  3a^-18  =  0  23.  17  «  + 31 +  2ar^  =  0. 

4.  2  =  5aj  +  7r^.  24.  18  -  41  a;  =  3  +  aj2. 

5.  6x-llx'=-7.  25.  10a;  +  25=5-2a;-ar^. 

6.  -51  +  42a;-3a^  =  0.  26.  3x  -  59 -^0^"=  0. 

7.  3a.-2  +  3a;  =  2a;  +  4.  27.  5x'  +  7x-6  =  0. 

8.  13-8a;  +  3ic2  =  0.  28.  ic*  +  12  =  7a;. 

9.  2iB2_,.iia5=32a;-x2_27.     29.  8a;-5ii-2  =  2. 

10.  176+3a;-ic2  =  2a;.  30.  5 a;  +  3ar^-22  =  0. 

11.  a^  +  e'a;- 54  =  0.  31.  50 -f  20  a;  +  a2  =  5  a;. 

12.  5  ar^- 9  a;  4- 12  =  4  ar^+ a;.      32.  a^  +  a; +  4  =  0. 

13.  2.^2 -4  a; -25  =  0.  33.  20  a;+2  ar'+42=33a;+a*. 

14.  7ar^  +  lla;  =  6.  34.  17  a;- 3ar»=  -  6. 

15.  2ar^-lla;  +  5  =  0.  35.  8a;  +  5ar^=-2. 

16.  2a;2-lla;  =  6.  36.  10  +  15  a; -f  a^  =  26  a;. 

17.  25a;-95  =  ar'.  37.  3ar^-2a;  -  7  =  0. 

18.  llar^-42a;  =  2.  38.  5  a^- 9  a;- 18  =  0. 

19.  ar2-8a;-4  =  a;-22.  39.  7  a;- 7  «-  + 24  =  0. 

20.  Sx^-\-5x=  -S.  40.  3lH-2a;  +  a^  =  0. 

41.  7a:2_|.7^_5,^^20  =  a;2_2a;  +  2. 

42.  5a;2  +  3a;-7  =  (a;-l)(a;  +  2). 

43.  (a;-3)2-(2a;-l)(2.r  +  l)+7  =  0. 

44.  3.T+(3a;-2)2  =  4a^-l. 

45.  9a^-(2a;-l)2  =  (a;-f3)l 

46.  7ar^  =  5a;-(a;-2)2  +  7. 

47.  (3a;-2)(3aj  +  2)  =  (2a;-8)^ 

48.  5a;-9a^  +  8(a;-x2>)^4^ 


206 


QUADRATIC   EQUATIONS 


DRILL  EXERCISES 

1.  Divide  y?  -\-  x^y  +  ^y^  +  ^2/^  +  xy^  +  y^  hy  x  +  y. 

2.  ])ivide  x^  —  y^  hy  :>?  -\-  xy  -f  y"^. 

3.  Divide  «^  —  ^/^  by  ar^  +  x^y  +  xy^  +  2/^. 

4.  Divide  x^  +  icy  -|-  y^  by  a;^  +  2/^. 

Find  the  square  roots  of : 

5.  x^  +  2x^-\-2a?  +  x'  +  2x-\-l. 

6.  a;8H-2«6_|_3a;4  +  2aj2_|_i.  ^ 

7.  a:^  +  4  a;5  +  10  x^  +  16  a^  +  17  a^  +  12  a;  +  4. 

8.  4a^-12a3-7a2  +  24a  +  16. 
Eationalize  the  denominators  of  the  following : 


9. 


10. 


15. 


16. 


19. 


V7- V4 
V7+  V2 
V7-V2' 


11. 


12. 


a-\-h 


V8  +  7 


x-^ay  =  c, 

bx-\-y  =  d. 

aa;  +  32/  =  2c, 

bx-2y  =  Sd. 

X     y 

1  +  1  =  6,     20 

y    z 

ax- 
[cx-\-dy  =  e. 


1  ,  1 

Z       X 


rx-^y-^z  =  a, 
2x-2y-{-2z=b, 
.3x  —  y  —  z  =  c. 


21. 


mx—y+bz=a, 
x  +  ay  —  z=l, 
^bx—y-{-az  =  b. 


22.  If  w  and  I  are  the  length  and  width  of  a  rectangle, 
express  in  symbols  the  length  of  its  diagonal. 

23.  The  lengths  of  the  two  sides  of  a  right  triangle  are  16 
and  24  respectively.  Express  the  length  of  the  hypotenuse 
in  the  simplest  form  without  approximating  a  square  root. 


SYSTEMS  INVOLVING  QUADRATICS  207 

SYSTEMS  INVOLVING  QUADRATICS 
201.   The  solution  of  two  equations  in  two  variables,  one  of 
which  is  linear  and  the  other  quadratic,  can  be  reduced  to  the 
solution  of  a  quadratic  equation  in  one  variable. 

Example.     Solve      (      x-\-y  =  3,  (1) 
\3x'-f  =  U.                                ,     (2) 

From  (1),                                  y=3-x.  (3) 
Substituting  in  (2)  and  reducing, 

2x'-\-6x-2S  =  0.  (4) 
Substituting  in  the  formula  §  199, 

^^-6±  v^36-4.2(-23)  ^-3±V55  .g. 

Hence  Xi  =  2.21  and  a^j  =  —  5.21. 

Substituting  these  values  of  x  in  (1)  we  have  as  the  approxi- 
^  =  2-211    ^^^.,=  -5.21 


=  2.21 1 
=  0.79  J 


2/1  =  0.79  J  2/2  =  8.21 

2/i  and  2/2  are  here  used  to  designate  the  values  of  y,  which 
correspond  to  a^  and  X2  respectively. 

EXERCISES 

In  the  above  manner  solve  the  following  systems  of  equations, 
finding  in  each  case  two  pairs  of  roots.  In  the  case  of  roots  which 
are  surds,  find  the  approximate  results  to  two  places  of  decimals. 

^     (x-y  =  l.  g      ra;-f42/  =  26. 

|ar'  +  2/'  =  13.  '    [x^-y'  =  ll. 

^     (x-\-y=9.  Arts.   x,  =  6,   y,  =  5. 

10.^  +  2/^  =  41.  a.,=  -9A,  2/.=  8H. 

(x-\-y  =  13. 
[xy  =  ^2. 

(3x—y  =  5.  6. 

y  +  y^^25. 


208 


QUADRATIC  EQUATIONS 


x  —  y  =  l. 
36     16     ^ 


5. 

5f  =  7, 


8.    P^  +  2/ 
•    I  0^-^3  2/' =  13. 


10. 


a;-32/=l. 
^ns.    flJi  =  4,   2/i  =  l- 


11. 


1 3  a;- 4  2/ =1. 


15. 


a;  -  2  2/  =  3.  . 

-471S.   rci=— 6.16, 2/i  =  — 4.68. 
a;2  =  .16,   2/2=  -1.42. 


16. 


17. 


X  -\-y  =  9. 

x''-2y^=-7. 

y  —  2x  =  5. 


y^  —  Sxy  =  16. 

Ans.   ici  =  3.71,  2/1  =  12.42. 
a^2  =  1.21,   2/2  =  2.58. 

r22/-3a^  =  0. 
'"•    12/^  +  0^  =  52. 


12     ra^  +  2/  =  4.  19     ly-2x  = 

l2a;2-3a;2/  +  2/'  =  8.  U'+2/'  = 


2/  —  2  ic  =  5. 
40. 


13. 


14. 


x-y  =  l. 
4:X^-\-2xy-y^  =  19. 


20. 


r  a;  -  4  2/  =  12. 

I3aj2  + 


f5a;4-S 
12x2- 


5x  +  y  =  12. 

3xy  +  f=0. 


2xy  —  6y  =  4A 
Ans.   a;i  =  4,   2/1  =  — 2. 

6 


0^2  =  -  If  2/2  =  -  3^1 . 


PROBLEMS 

In  each  problem  find  the  two  roots  of  the  quadratic  equation 
and  determine  whether  both  are  applicable  to  the  problem: 

1.  The  area  of  a  window  is  2016  square  inches  and  the 
perimeter  of  the  frame'  is  180  inches.  Find  the  dimensions  of 
the  window. 

2.  The  area  of  a  rectangular  city  block,  including  the  side- 
walk, is  19,200  square  yards.  The  length  of  the  sidewalk 
around  the  block  when  measured  on  the  side  next  the  street  is 
560  yards.     Find  the  dimensions  of  the  block. 


PROBLEMS  209 

3.  A  farmer  starts  to  plow  around  a  rectangular  field  which 
contains  48  acres.  The  length  of  the  first  furrow  around  the 
field  is  376  rods.     Find  the  dimensions  of  the  field. 

4.  A  rectangular  blackboard  contains  38  square  feet  and 
its  perimeter  is  27  feet.     Find  the  dimensions  of  the  board. 

5.  A  park  is  120  rods  long  and  80  rods  wide.  It  is  decided 
to  double  the  area  of  the  park,  still  keeping  it  rectangular,  by 
adding  strips  of  equal  width  to  one  end  and  one  side.  Find 
the  width  of  the  strips. 

6.  A  fancy  quilt  is  72  inches  long  and  56  inches  wide.  It 
is  decided  to  increase  its  area  10  square  feet  by  adding  a  border. 
Find  the  width  of  the  border. 

7.  A  city  block  is  400  by  480  feet  when  measured  to  the 
outer  edge  of  the  sidewalk.  At  4  cents  per  square  foot  it  costs 
$  416.64  to  lay  a  sidewalk  around  the  block.  Find  the  width 
of  the  walk. 

8.  A  farmer  starts  cutting  grain  around  a  field  120  rods 
long  and  70  rods  wide.  How  wide  a  strip  must  be  cut  to  make 
12  acres? 

9.  The  s^des  of  a  right  triangle  are  6  and  8  inches  respec- 
tively. How  much  must  be  added  to  each  side  so  as  to  increase 
the  hypotenuse  10  inches,  it  being  understood  that  each  side  is 
increased  by  the  same  amount? 

10.  A  rectangular  lot  is  16  by  12  rods.  How  wide  a  strip 
must  be  added  to  one  end  and  one  side  to  obtain  a  rectangular 
lot  whose  diagonal  is  1  rod  greater? 

11.  A  picture  is  15  inches  by  20  inches.  How  wide  a  frame 
must  be  added  to  increase  the  diagonal  3  inches  ?         ^ 

12.  An  athletic  field  is  800  feet  long  and  600  feet  wide. 
The  field  is  to  be  extended  by  the  same  amount  in  length  and 
width  so  that  the  longest  possible  straight  course  (the  diagonal) 
shall  be  increased  by  100  feet.  How  much  is  the  field  extended 
in  each  direction  ?     Ans.  71.36  feet. 


210 


QUADRATIC  EQUATIONS 


13.  A  and  B  start  from  a  certain  cross  roads  at  the  same 
time,  A  going  north  4  miles  per  hour  and  B  going  east  3  miles 
per  hour.  In  how  many  hours  will  they  be  16  miles  apart, 
measuring  in  a  straight  line  across  country  ? 

Let  t  equal  the  required  number  of  hours. 
Then         (4  0^  +  (3  0'^  =  16^  =  256. 
lQt^-\-9t^  =  256. 
25  <2  =  256. 
5  ^  =  ±  16. 
t  =  ±  3i 
The   solution    ^  =  —  3^   may  be    interpreted    as 
meaning  that  if  the    two    men   were,  traveling 
along  these  roads  in  the  same  direction  before  reaching  the  cross  road, 
they  would  be  1 6  miles  apart  3^  hours  before  meeting. 

14.  In  the  preceding  problem  if  A  goes  5  miles  per  hour  and 
B  4  miles  per  hour,  in  how  many  hours  will  they  be  24  miles 
apart  ?     Ans.  3.75. 

15.  A  rectangle  is  12  inches  wide  and  16  inches  long.  How 
much  must  be  added  to  the  length  to  increase  the  diagonal  4 

inches  ? 

_  ^  ^1 


Let  X  =  number  of  inches  to  be  added  to  the  length.  The  diagonal 
of  the  original  rectangle  is  V122  +  16^  =  20.  Hence  the  diagonal  of 
the  required  rectangle  is  24. 

Then  122  +  (ig  +  ^y  =  242, 

or  x2  +  32  a:  -  176  =  0. 

Solving,  xi  =  -  16  +  12V3"=  4.78, 

and  a:,  =  -  16  -  12\/3  =  -  36.78. 


PROBLEMS  211 

The  negative  solution  obtained  here  may  be  taken  to  mean  that  if 
the  rectangle  is  extended  in  the  opposite  direction  from  the  fixed 
corner,  we  shall  get  a  rectangle  which  has  the  required  diagonal. 
See  the  figure. 

16.  How  much  must  the  width  of  the  rectangle  in  problem 
15  be  extended  so  as  to  increase  the  diagonal  by  4  ? 

17.  A  trunk  30  inches  long  is  just  large  enough  to  permit 
an  umbrella  36  inches  long  to  lie  diagonally  on  the  bottom. 
How  much  must  the  length  of  the  trunk  be  increased  if  it  is  to 
accommodate  a  gun  4  inches  longer  than  the  umbrella  ? 

18.  A  rectangle  is  21  inches  long  and  20  inches  wide.  The 
length  of  the  rectangle  is  decreased  twice  as  much  as  the  width, 
thereby  decreasing  the  length  of  the  diagonal  4  inches.  Find 
the  dimensions  of  the  new  rectangle. 

19.  In  a  rectangular  table  cover  24  by  30  inches  there  are  two 
strips  of  drawn  work  of  equal  width  running  at  right  angles 
through  the  center  of  the  piece.  What  is  the  width  of  these 
strips  if  the  drawn  work  covers  one  tenth  of  the  whole  piece  ? 

20.  A  certain  university  campus  is  100  rods  long  and  80 
rods  wide.  There  are  two  driveways  running  through  the 
center  of  the  campus  at  right  angles  to  each  other  and  parallel 
to  the  sides.  What  is  the  width  of  these  driveways  if  their 
combined  area  is  356  square  rods  ? 

21.  A  farm  is  320  rods  long  and  280  rods  wide.  There  is  a 
road  2  rods  wide  running  around  the  boundary  of  the  farm  and 
lying  entirely  within  it.  There  is  also  a  road  2  rods  wide 
running  across  the  farm  parallel  to  the  ends.  What  is  the  area 
of  the  farm  exclusive  of  the  roads  ? 

22.  A  rectangular  park  is  480  rods  long  and  360  rods  wide. 
A  walk  is  laid  out  completely  around  the  park,  and  a  drive 
through  the  length  of  the  park  parallel  to  the  sides.  What  is 
the  width  of  the  walk  if  the  drive  is  3  times  as  wide  as  the 
walk  and  the  combined  area  of  the  walk  and  the  drive  is 
3110  square  rods  ? 


212  QUADRATIC  EQUATIONS       ' 

23.  The  Sum  of  the  sides  of  a  right  triangle  is  18  and  the 
length  of  the  hypotenuse  is  16.     Find  the  length  of  each  side. 

24.  The  length  of  a  fence  around  a  rectangular  athletic  field 
is  1400  feet,  and  the  longest  straight  track  possible  on  the  field 
is  500  feet.     Find  the  dimensions  of  the  field. 

Using  100  feet  for  the  unit  of  measure,  the  equations  are 

[x^  +  y^  =  25. 

25.  The  difference  between  the  sides  of  a  right  triangle  is  8 
and  the  hypotenuse  is  42.     Find  the  lengths  of  the  sides. 

'  26.  A  room  is  5  feet  longer  than  it  is  wide,  and  the  distance 
between  two  opposite  corners  is  25  feet.  Find  the  length  and 
width  of  the  room. 

27.  One  side  of  a  right  triangle  is  8  feet,  and  the  hypotenuse 
is  2  feet  more  than  twice  the  other  side.  Find  the  length  of 
the  hypotenuse  and  of  the  remaining   side. 

28.  A  vacant  corner  lot  has  a  50-foot  frontage  on  one  street. 
What  is  the  frontage  on  the  other  street  if  the  distance  between 
opposite  corners  along  the  diagonal  is  110  feet  less  than  twice 
this  frontage. 

29.  The  sum  of  the  squares  of  two  consecutive  integers  is 
13,945.     Find  the  numbers. 

30.  The  product  of  two  consecutive  integers  is  4422.  Find 
the  numbers. 

31.  A  square  piece  of  tin  is  made  into  an  open  box,  contain- 
ing 864  cubic  inches,  by  cutting  out  a  6-inch  square  from  each 
corner  of  the  tin  and  then  turning  up  the  sides.  Find  the 
dimensions  of  the  original  piece  of  tin. 

32.  A  rectangular  piece  of  tin  is  8  inches  longer  than  it  is 
wide.  By  cutting  out  a  7-inch  square  from  each  corner  and 
turning  up  the  sides,  an  open  box  containing  1260  cubic  inches 
is  formed.     Find  the  dimensions  of  the  original  piece  of  tin. 


REVIEW  QUESTIONS  213 

33.  By  cutting  out  a  square  8  inches  on  a  side  from  each 
corner  of  a  sheet  of  metal  and  turning  up  the  sides,  we  obtain 
an  open  box  such  that  the  area  of  the  sides  and  ends  is  4  times 
the  area  of  the  bottom.  Find  Ihe  dimensions  of  the  original 
sheet  if  it  is  twice  as  long  as  it  is  wide.  Ans.  41.48  in.  by 
20.74  in. 

34.  An  open  box  whose  bottom  is  a  square  has  a  lateral 
area  which  is  400  square  inches  more  than  the  area  of  the 
bottom.  Find  the  other  dimensions  of  the  box  if  it  is  10 
inches  high.  (By  lateral  area  is  meant  the  sum  of  the  areas 
of  the  four  sides.) 

35.  A  box  whose  bottom  is  4  times  as  long  as  it  is  wide 
has  a  lateral  area  600  square  inches  less  than  4  times  the 
area  of  the  bottom.  Find  the  dimensions  of  the  bottom  if  the 
box  is  6  inches  high. 

REVIEW  QUESTIONS 

1.  Explain  Ihe  method  of  solving  a  quadratic  equation  by 
factoring.  Can  you  apply  this  method  to  solve  the  equation 
(x  +  l)(x-2)=z5?     Explain. 

2.  Explain  the  method  of  solving  a  quadratic  equation  by 
completing  the  square.  What  is  the  Hindu  method  and  what 
are  its  advantages  ? 

3.  How  many  roots  has  a  quadratic  equation?  Find  the 
roots  otay^-{-2x-\-5  =  0.  Are  these  roots  in  the  form  of  any 
numbers  thus  far  studied?  What  are  such  roots  called? 
Solve  ar'  +  2a;  —  5  =  0.     What  are  these  roots  called ? 

4.  How  are  the  roots  of  the  equation  a^-^px  +  q  =  0  related 
to  p  and  q?     How  may  this  be  used  in  checking  the  solutions  ? 

6.  Do  both  roots  of  a  quadratic  equation  necessarily  satisfy 
the  conditions  of  the  problem  from  which  such  an  equation 
may  be  derived?  In  checking  the  solution  of  a  problem  is 
it  sufficient  to  make  the  test  alone  in  the  equation  derived 
from  th^  problem? 


214 


QUADRATIC  EQUATIONS 


DRILL  EXERCISES 

Simplify  each  of  the  following  as  much  as  possible  without 
approximating  roots  : 

1.  V32  4-V72-V50. 

2.  2a-VaFb-b^¥a-{-Vab. 

3.  Va«  +  2a:'b-j-  ab^  +  Va'  -2  a'b  +  ab^  -  2  Va^. 

4.  Va'  -  d'b  -  Vft'a  -  6"=  +  Va«&'  -  a^ft^. 


5.    V(a;  +  2/)(a^'  -  /)  -  V(a;  +  2/)X^  —  2/)  -  Va^  -  ar^?/. 
Solve  the  following  equations  : 


6.  ■Vx'-S  +  x^S. 

7.  V5a-24  +  4  =  V5a. 


8.    V2a-l  =  7-V2a  +  6. 


9.    Va;+2=Vi«— 6  4-2Va;— 5. 
Rationalize  the  denominators  of  the  following : 


10. 


11. 


■\/a  +  ^x 
Va  — V^ 

Va  +  a;  -f-  Va 


Va  +  a;  — Va 
Solve : 

62/ =  2, 
3  62/  =  6. 

x  —  y  —  Sz=  —  6, 
2x-\-y-z  =  ll, 
[-^x-\-Sy  +  z  =  16. 


14 


16. 


{3  ax 
2a;  +  ' 


12. 


13. 


15. 


17. 


a  — V6 


V  a;  —  a  +  6 


Va;  +  a  —  c 

2  a;  —  3  fey  =  c, 
2ax  —  5y  =  d. 

'2x-y  +Sz=20y 
x-^4ry  —  z=  —  2, 
5x-i-y  —  6z  =  6. 


18.  Divide  a^ -Sx*-lHa^ -{-24.x^  +  52x-21  by  a^^+aj-j. 

19.  Divide  6  a'-\-5  a^-60a^-\-4:a'-\-71  a+28  by  3  a^-5  a-4. 

Find  the  square  roots  of :  • 

20.  IQx^-^Oxy  +  x^f-^-SOx'y^-^dfK 


21.   4m«-20m*  +  41m^  +  52m3-14 


m' 


24m  +  9. 


CHAPTER   XIII 

ALGEBRAIC  FRACTIONS 

COMMON  FACTORS 

Simple  algebraic  fractions  have  already  been  studied  in  Chap- 
ter V,  where  they  were  treated  exactly  as  fractions  in  arith- 
metic. This  was  sufficient  for  all  our  purposes  up  to  this  point. 
We  now  take  up  a  more  formal  study  of  fractions. 

202.  If  a  number  is  a  factor  of  each  of  two  or  more  numbers, 
it  is  said  to  be  a  common  factor  of  these  numbers. 

Thus,  8  is  a  common  factor  of  16  and  48,  and  12  is  a  common 
factor  of  12,  36,  and  48. 

If  each  of  a  given  set  of  numbers  is  separated  into  prime 
factors,  any  common  factor  which  they  may  have  is  at  once 
apparent. 

Illustrative  Example.     Find  the  common  factors  of 

10(x  +  y)\x  -  y),  5(x  +  y){^-  y%  and  15(x  +y){o^-  /). 

Factoring,  10(a:  +  y)\x  -  y)  =  2  •  5(ar+  y)(a:+  y){x  -  y). 
b{x-\-y){x^-y^)  =  5(x  +  y){x  +  y)(x-y). 
15(x  +  y) (x«  -  1/8)  =  3  .  5(x  +y)(x-  y){x^  ^  xy -\-  y^). 

The  common  prime  factors  are  5,  aj-f-i/,  and  x—y.  The 
other  common  factors,  obtained  by  combining  these,  are  5(a;-f-?/), 
5(x—y)  and  5{x  -\-y){x  —  y).  The  last  factor,  5(a; -|- y){x  —  2^)  is 
called  the  highest  common  factor. 

The  name  highest  instead  of  greatest  is  used  in  algebra  referring 
to  the  number  of  prime  factors  which  enter  into  it.  Thus,  x^  is  of 
higher  degree  than  x,  although  \i  x—\,  x^  is  not  greater  than  x. 

203.  Definition.  The  product  of  all  the  common  prime 
factors  is  called  the  highest  common  factor.  This  is  usually 
abbreviated  to  H.  C.  F. 

215 


216  ALGEBRAIC   FRACTIONS 

EXERCISES 

Find  the  H.  C.  F.  of  the  following  sets  of  expressions  : 

1.  x  —  y,ix?-y'^,Q(?-2xy'\-y\ 

2.  a?  +  2x-{-l,Zx  +  Qx''-\-Zx'. 

3.  a!2  +  4a;  +  4,  a^-6a;-16. 

4.  a^-8a:  +  16,  a;2  +  10a;-56. 

5.  a^-h^,  a^-2ab  +  b\ 

6.  a^-\-f,x'-y%x'-^2xy  +  y\ 

7.  a^-7x-{-12,ax-~Sa-bx  +  Sb. 

8.  a^-13a  +  4.2,a^-216,d'-a-S0. 

10.  b^  +  7b-  30,  &2  +  11  6  -  42,  62  _  5  _  g. 

11.  a^  +  2a^-^a,a^-\-a,a^  +  5a^-{-4:a. 

12.  0^  +  2/^,  aJ3  4-a;V  +  a;/  +  2/^• 
13.  ic^-f  3a^4-2a^,  a^  +  a^,  a;^+7a^4-6ar^. 

14.  aj2  — lla;-l-30,  JC2;  — 5;2  +  if2  — 5a;. 

15.  m^  —  n^,  2  a^m^  +  2  «^mn  -|-  2  ar^?i^ 

16.  a;2-l,  ar»-l,  a;2-13a;+12. 

17.  l-64a^,  l-16a^,  5-22-20a;  +  8a;2J. 

18.  l  +  125a^l  +  10a  +  25a2,  l-25a2. 

19.  ac-'ax  +  Sbc  —  Sbx,a^-\-27b\ 

20.  5c-2,  5ac  +  20c-2a-8. 

21.  4:Ccl^-x^,2ai^-{-c^-ar^,2x*-Sa^  +  x^. 

22.  3a3-3a,  3a3-6a2  +  3a,  6a«  +  12a'-15a. 

23.  6x—10xy  +  4:  xy\  18  a;  —  8  xy"^,  54  a;  — 16  x'l^. 

24.  3 aj«  +  9 a;^- 3a;3,  5 x'y''  +  15  xy'' -5 /,  7  0x^+21  ax-7 a. 

25.  18  a^  -  57  a;2  _^  3Q  3,  9  ^  _  ^5  ^2  _^  g  3,^  ^g  ^.39  ^j^^^  y. 


COMMON  MULTIPLES  217 

COMMON  MULTIPLES 

204.  A  number  is  said  to  be  a  multiple  of  any  of  its  factors. 

In  particular  any  number  is  a  multiple  of  itself  and  of  one. 

Thus,  18  is  a  multiple  of  1,  2,  3,  6,  9,  and  18,  but  not  of  12.  3  a^x^ 
is  a  multiple  of  3,  3  ar,  3  x^,  etc. 

Since  a  multiple  of  a  number  is  divisible  by  that  number,  it 
must  contain  as  a  f actoi'  every  factor  of  that  number. 

E.g.,  108  is  a  multiple  of  54  and  contains  as  factors  all  the  factors 
of  54,  namely  3,  3,  3,  and  2,  and  also  2,  6,  9,  18,  and  54. 

Definition.  A  number  is  a  common  multiple  of  two  or  more 
numbers  if  it  is  a  multiple  of  each  of  them. 

Thus,  18  is  a  common  multiple  of  6,  9,  and  18.  Evidently  3  •  18, 
4  .  18,  5  •  18,  etc.  are  also  common  multiples  of  6,  9,  and  18.  Of  all 
these  common  multiples  18  is  called  the  lowest  common  multiple. 

205.  The  process  of  finding  the  lowest  common  multiple  of 
a  set  of  expressions  is  shown  as  follows : 

Illustrative  Example.     Find  the  lowest  common  multiple  of 

x^  -  y^;  x2  +  2  a:?/  +  2/2.  and  x^  -  2  xy  -\-  y\ 
Factoring,  x'^  -  y^  =^  {x  -  y)  {x  ■\- y) .  (1) 

a;2  +  2  xy  +  y2  =  (a:  +  y){x  +  y).  (2) 

x^-2xy-^ry^={x-y){x-y).  (3) 

In  order  that  an  expression  may  be  a  multiple  of  (1)  it  must  con- 
tain the  factors  x  —  y  and  x-\-  y.  To  be  a  common  multiple  of  (1) 
and  (2)  it  must  contain  an  additional  factor  x  -\-  y,  that  is,  it  must  con- 
tain (x  —  y),  (x  +  y),  (x  +  y).  To  be  a  common  multiple  of  (1),  (2), 
and  (3)  it  must  contain  an  additional  factor  x  —  y,  that  is,  it  must 
contain  (x  —  y),  (x  +  y),  (x  -{-  y),  {x  —  y).  The  product  (x  —  y) 
(x  +  y)(x  4-  y)(x  —  y)  =  (z  —  yy(x  +  yY  is  called  the  lowest  com- 
mon multiple  of  (1),  (2),  and  (3),  since  it  is  the  common  multiple 
which  contains  the  smallest  number  of  prime  factors. 


218  ALGEBRAIC  FRACTIONS 

In  general,  the  process  may  be  described  as  follows :  to 
obtain  the  lowest  common  multiple  of  a  set  of  expressions,  factor 
each  expression  into  prime  factors  ;  use  all  factors  of  the  first 
expression  together  with  those  factors  of  the  second  which  are  not  in 
the  first,  those  of  the  third  which  are  not  in  the  first  and  second,  etc. 

It  is  evident  that  in  this  manner  we  obtain  a  product  which 
is  a  common  multiple  of  the  given  expressions,  but  such  that 
if  any  one  of  these  factors  is  omitted,  it  will  cease  to  be  a  mul- 
tiple of  some  one  of  the  expressions ;  that  is,  it  will  no  longer 
be  a  common  multiple  of  them  all. 

Thus,  if  in  the  example  above  either  of  the  factors  x  —  y  i^  omitted, 
the  product  will  no  longer  be  a  multiple  oi  x^  —  2xy  -\-  y\ 

206.  Definition.  The  lowest  common  multiple  of  a  set  of  ex- 
pressions is  that  common  multiple  which  contains  the  smallest 
number  of  prime  factors.  The  lowest  common  multiple  is 
usually  abbreviated  to  L.  C.  M. 

EXERCISES 

Find  the  L.  C.  M.  of  the  following  expressions : 

1.  2-3.4;  3.7.8;  23.3-4. 

2.  5  ccy,  10  :i(?y,  25  a?y. 

3.  2  ah,  6  a^,  4  hh. 

4i.  x^  —  y^,  a^  —  2  xy  -\-  y\ 

5.  x-y,x-\-y,  x?  —  y^. 

^.  ^-^,2-x,2-\-x. 

7.  a^A-2ah^-h'',a^-2ah-\-h\ 

8.  0^2  4-  3  a;  -h  2,  i«2  _  4^  3,2  _  ^ 

,     9.  25a^-l,  125a^-l. 

10.  2x^-7x  +  6,4:x'^-llx-\-6. 

11.  x^  —  f,Xr-y,x'-{-xy  +  y\ 

12.  a^  —  y^,3(^-\-y^fX^  —  'if, 


REDUCTION  OF  FRACTIONS   TO  LOWEST  TERMS     219 

13.  5a^  +  7x-6,a^-15x-M. 

'  14.  x^-^f,x^  -y^,  (x  —  yf. 

15.  3  a6c,  a^  — 4ac-f4c^,  a— 2  c. 

16.  a;'^  —  1,  a;  4- 1,  0^  H- 8  a;  +  7- 

17.  4a^i/-44a^y  +  120a^,  3aV-22a3a;  +  35a^ 

18.  x'+2xy-\-f,2ax^-10ax-{-12a. 

19.  36x2-21  6a;  +  366,ic2-5ar  +  4. 

20.  5a262_6ttV,62^_26c  +  6H-c+c2. 

21.  15  c^oa;"  +  16  c^aa;  +  c^a,  2  caar^  + 10  caa;  +  8  ca, 

REDUCTION  OF  FRACTIONS  TO  LOWEST  TERMS 
.    207.    By  Principle  XV  any  factor  common  to  the  numerator 
and  denominator  of  .a  fraction  may  be  cancelled.     That  is, 

ak  _a 

bk~  b' 

3 
Thus,  2.3'4.5^2>4.5.  "^^'^'^-jx^  3a:  ^3a;. 

3-7 -11        7    11  '    ;2*.3.^2      2.4      8  ' 
2  4 

x3  -  7a;  +  12  ^  {i ^jx  -  4)  ^  x  -  4 

a;2_6x  +  6       (a:-2)(*^-^)      x  -  2' 

If  the  terms  of  a  fraction  have  no  common  factor,  the  frac- 
tion is  said  to  be  in  its  lowest  terms. 

EXERCISES 

Reduce  the  following  fractions  to  lowest  terms : 


1. 


2*.  53.  9^        '   a^b^ 

4a*6^V  ^^2xv  + 

o. 


y?fz\  g    ar^  +  7  a;  _  30 

xy^T^  '    Qi?  —  lx +  12      "    xy  —  5x  +  Syz  —  15z 


2a^- 

-Sxy  +  y^ 

64 

:-6- 

16- 

86  + 6^ 

a:«4-27  2» 

220  ALGEBRAIC  FRACTIONS 

l-216c^  19    4fl;^-28a^H-48a^ 

14.hz-2hx  +  ax-laz  9  a^6^  +  18  a^6^c  +  9  a^6^c^ 

j2         3a2-29a  +  56  ^^     7  x/ -  133  a;y  +  126  a; 

'    63  —  9  a  —  7  m  +  ma '  *     15  a;?/^  —  36  ic?/  +  21  a; 

13  «(a^-yy  22    20a^  +  20a^2/  +  5a;y^ 

*  (x'-f)(x-y)'  '         60a^-15a^f 

14  a:^+27  3a6^-3«6V 
4«2^24xH-36'                     '   27a36^  +  27a36c* 

a2-3a-36  +  «&  04    4a^-42a^  +  20a 

*  (a2-62)(a_3)    '  *      2  a'b^  -  20  d'b^ 

^g    2^  •  3^  •  5^  -  2^ .  3^ .  5^         ^g     (a;-l)(a;-2)(a;-3)(a;-4) 
2*.32.5^-25.32.5'^*       ^*    (a;-l)(a;-3)(a;-3)(a;-4)  ' 

6aj^/  +  3a^/         *         *     {a^ -  2 xy -h  y'){x  +  y) 

5c  +  10b-bc-2b'  (a.-^-l)(a^  +  l)(3a^2  +  3) 

8c3  +  6463  3(a;^-l) 

REDUCTION  OF  FRACTIONS  TO  A  COMMON  DENOMINATOR 

208.   By  the  formula  f  =  77  any  factor  may  be  introduced  into 
0     ok 

the  numerator  and  denominator  of  a  fraction. 

In  this  manner  any  fraction  may  be  changed  into  an  equal 
fraction  whose  denominator  is  any  given  multiple  of  the  de- 
nominator of  the  given  fraction. 

'^'  4      4.5'    a +  6      la  +  b)(a-hb)       (a +  6)2* 

Any  two  or  more  fractions  may  therefore  be  changed  into 
respectively  equal  fractions  which  shall  have  a  common  denomi- 
nator, namely,  a  common  multiple  of  the  denominators  of  the 
given  fractions. 


Thus, 


REDUCTION  TO  A  COMMON  DENOMINATOR  221 

Illustrative  Example.     Reduce  ^^,^-±i,  2x-{-S  ^^  ^^^^_ 

x-i-1    x  —  1     x^  —  1 
tions  having  a  common  denominator. 

The  L.  C.  M.  of  the  denominators  is  (x  —  l)(x  +  1).  Multiply  the 
numerator  and  denominator  of  each  fraction  by  an  expression  which 
will  make  the  denominator  of  each  new  fraction  (x  —  l)(x  -\-  1). 

x-l^(x-l)(x-l)  ^     a:g-2a;  +  l 
x+1      (x+l)(2:-l)       (x-hl){x-iy 

x+1  ^  (x+l)(x+l)  ^        (x  +  ly       . 
x-1      (^x-l){x  +  lj      (x  +  lXx-1)* 

2x  +  3  ^         2a;  +  3 
x^-1  ~(a;  +  l)(a:-l)* 

It  is  best  to  indicate  the  multiplication  in  the  common  denomina- 
tor, since  this  makes  it  more  easily  apparent  by  what  expression  the 
numerator  and  denominator  of  a  fraction  must  be  multiplied  in  order 
to  reduce  it  to  a  fraction  with  the  required  denominator. 

209.  The  Three  Signs  of  a  Fraction.  It  should  be  noticed  that 
there  are  three  signs  in  connection  with  a  fraction:  the  sign 
of  the  fraction  itself,  the  sign  of  the  numerator,  and  the  sign 
of  the  denominator.  Any  two  of  these  signs  may  be  changed 
simultaneously  without  changing  the  value  of  the  fraction. 

Thus,    (1)|=5--.      (2)|  =  -f^.      (3)f  =  -^"- 

Show  why  each  of  these  statements  is  true.  What  princi- 
ples are  involved  in  each  case? 

How  may  the  sign  of  the  numerator  of  a  fraction  be 
changed  if  it  is  in  the  form  of  a  polynomial  ?  If  it  is  in  the 
form  of  a  product  of  several  factors?  Take  care  to  distin- 
guish these  two  cases. 

This  is  useful  in  cases  like  the  following : 

X  A-X         X  1 

Reduce  -^ >  — — ->  and to  fractions  having  a  com- 

1-x    x^-1  x  +  1  ^ 

men  denominator. 


222  ALGEBRAIC   FRACTIONS 

l-x       x-1  (a;  +  l)(a:-l)  x^  -  1        ' 

^     ,and      1  ^-^  ^-1 


a;2_i      x2-l  x  +  l      (a;  +  l)(x-l)      a;2  - 1 

Apply  the  sign  changes  shown  in  (1),  (2),  and  (3)  above  to 
each  of  the  following  fractions  : 

(^)^c'     ^'^'^a^     ''^^a^     (4)(^z:^)^ 


(l-a;)(a;  +  2)'     '"        d-c'     ^'        {d-a)(b-a) 


EXERCISES 

Reduce  each  of  the  following  sets  of  fractions  to  equivalent 
fractions  having  a  common  denominator. 

x±S  4  4    _2_    a;-l    a^  +  l. 

*  x-y'  x'-'Zxy  +  i/  '   3  -  a;'  x  +  l'  a:-3' 


3          ^-^                 ^+4  g              1                        1 

a:2_9a;+20'  7ar^-26a;-8  *   a;2-3  a;-4'  a^  +  3  a;  +  2* 

^  g+l a-1  g         1        _1 1 

a^-2ab  +  b^'  0^+2  ab-^b^'  '  a^ -W'  b  -  a  a'-^-ab^W 

^                a  be 


5a--4:a-12'  a^-\-4.a-12'  a-2 
10  ^-^  a;4-2  a;-l 

11.     ^^        ^  12.  ^^'  1         13       g         6 

m— l'  l  +  n  (E-{-r){7n—iy  B+r  '  n—a  n  —  b 


REDUCTION  TO  A   COMMON   DENOMINATOR  223 


14  ^(T-  Q)     F  16     -^^    _i ^. 

w{Q-t)'w  '  a-A'  E~\-r   A-a 

15  ^  F  1  j^    ^    r    _J^       jga? 

F— v'  F+v'  F^-*2;2  «'  y'  x  —  y^  ^  +  y 

210.  Since  any  number  may  be  written  as  a  fraction  with 
the  denominator  1,  the  above  process  may  be  used  to  reduce 
an  integral  expression  to  the  form  of  a  fraction  having  any 
desired  denominator. 

Thus.  3  =  5-^^  X  -  3,  =  (^-.VX^^-I),  etc. 

0  x-*  —  1 

It  is  sometimes  convenient  to  reduce  expressions,  some  of 
which  are  not  fractions,  to  the  form  of  fractions  having  a  com- 
mon denominator. 

Illustrative  Example.     Reduce  5  x,  ,     ^~^,  to  frac- 

ar  — 1      x  —  1 

tions  having  a  common  denominator.     The   lowest   common 

denominator  is  ar'  —  1. 

Thus,        5^3,5x(x^- 1)^5x8-53:     5x-1^5a:-l 

2  a;  -  y  ^  (2  a;  -  y)(x  +  1)  _  2  x^  +  2  a:  -  yx  -  y. 
x-1         (a;-l)(x+l)  x^-l 


EXERCISES 

Reduce  the  following  expressions  to  fractions  having  a  com- 
mon denominator : 

ar^4-2a;^  +  2/^  aJ-y 


,  2c4-2.  5.   a^-a^  +  2/^^-2/^ 


a;-?/  '  a;  +  ?/  '  '  aj  +  y 


!4 

ALGEBRAIC  FRACTIONS 

6. 

x-y   x  +  y 

7. 

a  —  o   b  —  c 

8. 

a^     1    «»     1  ^  +  ^. 
'            '«-l 

9. 

^  +  2»^  +  ^'4,'l'.- 

10. 

x^  +  y^   x  —  y 

n.     ^\,r. 
a  — A 

12.      -^^.i?      n            13.    2',«  +  ' 
ft  —  ^                                        2 

14.     ^(g-0,  . 
w 

:r,£^.              15.^/^^. 

ADDITION  AND  SUBTRACTION  OF  FRACTIONS 
211.   Bj  Principle  III  read  in  the  reverse  order : 

o     c        c  c     c        c 

If  fractions  which  are  to  be  added  or  subtracted  do  not  have 
a  common  denominator,  they  must  be  reduced  to  this  form. 

Example.    Add    «^  and  S!L±1^^+^. 
a  +  &  a2  -  2  a6  +  62 

Reducing  the  fractions,  to  the  common  denominator 
we  have  («-*)(«-*)«.  + *), 

a-h  ^{a~h){a-  b)(a  -b)  ^a^-^a^  +  S  ah^  -  ¥ 
a  +  b      (a  +  b)(a  -  b)(a  -  b)      (a  +  b)(a  -  b)(a  -  6)' 

and  Q^  +  2  q&  +  6^  ^  (a  +  b)  (a^  +  2  ab  +  b^)  ^  q8  +  3  a^5  +  3  a&2  +  &« 
a2  -  2  aft  +  63       (a  +  6)(a  -  6)(a  -  6)       (a  +  b){a  -b){a-b)' 

Adding  the  numerators,  we  have  2  a^  +  6  aft^ ;  whence  the  sum  of 
the  fractions  is  2  «»  +  6  aft^ 

(a  +  6)(a-6)(a-6)' 


ADDITION  AND  SUBTRACTION  OF  FRACTIONS       225 

EXERCISES 

Perform  the  following  additions  and  subtractions : 


1. 

3      2       a  +  b 
4:'^7     Sa-b 

2. 

x—y        x—y 

3 

a^-9a;  +  18          x 

ar^-13a;  +  36  '  4-a; 

4. 

S     a+b      a-b 

5. 

3            5            2 

23 .  32     22 .  3*     2* .  33 

a 

a2_9&2          a'-Gab 

a^^6b  +  9b'     a'-db^' 

7. 

x+y ^x-y     2 
x-y     x-\-y 

8. 

2a?           y             y     ^ 
Q(?-f     X-y      x->ry 

9. 

a  +  1             a-1 
a^  +  a  +  l     a^-a  +  l 

10. 

6            &             b^ 
1-b     1  +  b'     1-6^* 

11. 

ic  +  l      a;  +  l      3ic-|  2 

a;_2      a;  +  2  '    x'-4. 

12. 

x-1      x  +  1     x'-B 
x  +  1      x-1      x^-l 

13. 

y^    ,    y        y   . 

^-1  S  +  1     1-2/ 

14. 

11     1    I  -1 

a;     y     x  —  y     x  +  y 

15. 

1  &           2a2         -a 

2  b-a     W-a"     b^a 

226  ALGEBRAIC   FRACTIONS 


^g     a^-{-4.xy  ^       1 


Q?-\-y^       x-\-y     x^  —  xy-\-y^ 
17.    .-J-,+     1  ^ 


1  —  a^     1  —  x     1  +  x  +  a^ 

iQ  a  — 3  a—  1        ,  a 

^°'     —o 7. TT. o ;:; t~^-t 


a2_3a  +  2      a^-^a  +  &     a2-4aH-3 
19.    ^ ^ +     ^ 


20. 


x'-^x-lA:     x-7     ar^-9a;  +  14 

a b 

ac -\- ad  —  he  —  hd     a^  —  2ah  +  h^ 


21.    -!,+   „,  f      .     ,      3 


a  —  5      a2  4-3a  —  40      a  +  8 


23     ^ I ^ . 

a^  +  4a;-60      a^-4a;-12 


24.      ^-^  +-    ^-« 


25. 


a^  —  (?     c«^  +  2  ac  +  c^      a  —  c 

9 8 

a^  +  7ar_18      y^^^x-Vo" 


26.    — A±l__|_        <*-^  ^  +  2 


27. 


a^-a-e     a2-7a  +  12     a2-2a-8 

? i-  +  _l_. 

a;2-lla;  +  30     a^-36     a^-25 


28.     .        },     ...  +  .     .2        „.+ 


29. 


(a5_l)(a;  +  2)      (a;4-2)(a;-3)      (iB-l)(3-a;) 

1 1 + ^ 

(a-  6)  (6  _  c)      (6  -  a)  (c-  c?)      (5  -  c)  (c  -  (^) 


3Q     _4 g-l  ft2-38q-3 

■    a-3      a2  +  3a+9  oc' -21 


DRILL  EXERCISES  227 

DRILL  EXERCISES 

Find  the  square  root  of  each  of  the  following : 

1.  25  a^  H- c2  +  9  62  _  10  ac  +  30  a6- 6  6c. 

2.  a^-\- 2/^+42:^+4 v^—  2xy -\-Axz  —  4:XV  —  4:yz  +  4:yv  —  Szv. 

3.  x^-2x^-ai^  +  3x^-\-2x  +  l, 

4.  x'-6a^  +  13a^-12x  +  4:. 

5.  Divide  6x*-ha^-\-12x^-\-Shj  20? -x-{-2. 

6.  Divide  3a^  +  4ar^-a^  +  6a^-12ar^+8x-12bj3ic2-2a;+3. 

7.  Divide3a'-5a^  +  8a3  +  2a2-18a  +  12  by  a^-a-\-2. 

8.  Divide  6  a^  +  10  a«- 9  a^  +  ll  a-6  hy2a^  +  Aa-S. 

To  each  of  the  following  binomials  add  one  squared  term  so 
as  to  make  it  a  trinomial  square : 

9.   4a2  +  8a.  12.   25ar^-7a;.  15.   Sx^-4:X. 

10.  9a2  +  30a.  13.    7  6^-36.  16.   7x2-lla;. 

11.  x^  +  3x.  14.   166^-76.  17.   8ar^  +  7a?. 

Reduce  the  solution  of  each  of  the  following  to   simplest 
form  without  approximating  any  roots: 

18.  7a^=27.  24.    7  aar^  =  98. 

19.  5ar^=108.  25.   2(a  +  6)^2  ^  30o. 

20.  2ar^  =  3.  26.    a^:       ^^^ 


21.   3a2  =  343. 


27.    x^ 


125  ab 
128 


22.  5a^  =  Sa^b.  72  cd^ 

23.  2ar'  =  27  a(a  + 6)1  28.  (a -  b)x^  =  a  +  b. 

Solve  each  of  the  following  and  test  results  by  the  method 
given  in  §  198. 

29.  »2~7a;+9  =  0.  33.  ar^  +  26a;=3c. 

30.  2  i»2- 5  a; +  2  =  0.  34.  2x^-5ax  =  a^. 

31.  7ar^  +  18x-3  =  0.  35.  aaf  +  2bx  =  Sc. 

32.  a;2_i2a;  +  16  =  0.  36.  4  aV  +  6  6a;  =  3  6^. 


228  ALGEBRAIC  FRACTIONS 


MULTIPLICATION  AND  DIVISION  OF  FRACTIONS 

212.  Since  a  fraction  is  an  indicated  quotient,  and  since 
multiplying  the  dividend  or  dividing  the  divisor  multiplies 
the  quotient,  it  follows  that  the  product  of  a  fraction  and  an 
iyitegral  expression  is  obtained  by  multiplying  the  numerator  or 
dividing  the  denominator  by  the  integral  expression. 

Thus,  4.^=il^  =  ^  or4.^  = 


That  is,  in  general,  a 


8       8        2  8     8^4     2 

b     a*  b     ab        a 


c        c        c      c-i-b 


It  is  best  to  factor  completely  the  expressions  to  be  multi- 
plied and  to  keep  them  in  the  factored  form  until  all  possible 
cancellations  have  been  made^ 


EXERCISES 

Find  the  following  indicated  products  and  reduce  the  frac- 
tions to  the  simplest  form : 

1.   (l-a)xl±^.  -i^-^Xfi^- 

3.   (a?-2xa  +  a?) X  ^±-^.  9.  (a^  +ab+  b^ X  -|^,. 

^  '     x-a  '     a^  —  V 

*■    :#^4fiX(»^-ll'^+18)-   10.  (l_a  +  a^x^. 
ar— 5ic+6  a^  +  1 

a2_8a-fl6     ^  ^  x'-l      (aJ  +  1)' 

6.    (a^-f  9a;-t-l8)  X -^— 77 —.   12.         '   . .  X  — — -  • 

^  ^      ar^— 2aj— 15  a^  —  b^     a^  —  b^ 


MULTIPLICATION  AND  DIVISION   OF  FRACTIONS     229 

213.  Since  multiplying  the  divisor  or  dividing  the  dividend 
divides  the  quotient,  it  follows  that  a  frciction  is  divided  by  an 
integral  expression  by  dividing  the  numerator  or  multiplying  the 
denominator  by  the  integral  expression. 

Thus,  --.2  =  -^  =  -     or-.2=-x.-  =  ^-^=-. 

That  is,  in  general,   -  -^-  c  =  - —  =  -^ — 
0  0  •  c         0 

EXERCISES 

Find  the  following  indicated  quotients  and  reduce  the  frac- 
tions to  their  lowest  terms : 

x  —  y 
a^-2x  +  l      ^  ' 

xy-ix-Zy  +  Vi     ^  ' 

8.  »^  +  «^  +  6x  +  a6  ^(^  +  tu;_5a--5a). 
a^  +  aa;-3a;-36      ^ 

9.  mr  +  m«-nr-ns^.3^_^^3^_^^)^ 

ma;  —  m  —  ?i£c  -f  ?i 


230  ALGEBRAIC   FRACTIONS 

TO   MULTIPLY   A   FRACTION    BY    A   FRACTION 
214.    To  multiply  a  number  by  the  quotient  of  two  numbers 
is  the  same  as  to  multiply  by  the  dividend  and  then  divide  the 
product  by  the  divisor. 


Thus,  4.3      /4     3>|_^  2^4^2 

9     2      V9        y  3 


Ingeneral,     ^  .  ^=  ff  .  c") -^  t/  =  ^-^  (/:    ^^ 


b     d     \b  b  bd 


That  is,  the  product  of  two  algebraic  fractions  is  a  fraction 
whose  numerator  is  the  product  of  the  given  numerators  and 
whose  denominator  is  the  product  of  the  given  denominators. 

Illustrative  Example.  Multiply  „  ^"-^ —  by  ^-^^+^ 
and  reduce  the  resulting  fraction  to  its  lowest  terms. 

x^-1  ar^  -  3  a:  +  2  _  (x  -  l)(-3^~^)(»^^)(x  -  1) 


a:2  _  7  a;  +  10      a;^  +  2  a;  +  1      (t— 2-)  (x  -  5)(-xH~^)(x  +  1) 

=  (x-l)(x-  1)  ^  a:^  -  2  a:  +  1 
(a;  —  5)  (x  +  1)      a;2  —  4  a;  —  5 

It  is  desirable  to  resolve  each  numerator  and  denominator 
into  prime  factors,  and  then  cancel  all  common  factors  before 
performing  any  multiplication. 

EXERCISES 

Find  the  following  indicated  products  and  reduce  each  frac- 
tion to  its  lowest  terms : 

S^y     6^  12  c^b       35(c^+c&4-&^ 

'    2yz^      9a^'  ''  5(c^-b')  Uc'b' 

5a(a-b)       9(a-\-by  f^3y+2     y'-Ty+12 

'   3c(a  +  b)     15(a'-b^  '  y'-5y-^6      y'+Sy+7  ' 


TO  DIVIDE   A  FRACTION  BY  A  FRACTION  231 

5    3^ .  4«     10  . 2  ^-x      2^J-^x±2 

'    5'-2*        3*     *  '   x^-l         3ar^  +  6a; 

3^«     5^2      6^  g^-lOa  +  ie      a  +  3 

x^  +  a;?/  —  a;^      (a;  —  2)^  —  3/^      ^  —  y^  —  yz 

10         3(0^  +  4)^  (a: -7)^        , 

•  4(a;+'4)(a;-7)      3(a;  +  4)(a;-7) 

•  a2_7a-144      a(a-4)(a  +  2)* 

^2    3  a(a  +  7)(a  -  5)      b(a  +  3)(a  + 10) 
'  7  6(a  +  3)(aH-7)      a(a-5)(a- 10) 

3^-2f-l      2  ^'^  +  5^-3     4^'^  + 10  ^  +  4 

•  2t'  +  t-l       Zf  +  lt  +  2      4.^-21-2' 

^^     6a^-7x  +  2       6ar^-5a;-l     10x^4-3a;-l 
10ar^-7a;  +  l       Gx'  +  x-l        6ar^-4ar-l  * 

4:b'-nb  +  4.     Wb^-21b-\-9     Sb'-5b  +  2 
'    662-76  +  2      56^-236  +  12     46^-56  +  1' 


TO  DIVIDE  A  FRACTION  BY  A  FRACTION 

215.  To  divide  a  number  by  the  quotient  of  two  numbers 
is  the  same  as  to  divide  by  the  dividend  and  then  multiply 
the  result  by  the  divisor. 

Th..  |.|  =  (|..).3  =  |.3  =  |. 

In  genera,,  2.?  =  g.o)x.=  (^^)x.  =  '^. 

That  is,  a  number  is  divided  by  a  fraction  by  inverting  the 
fraction  and  multiplying  by  the  new  fraction  thus  obtained. 


232  ALGEBRAIC  FRACTIONS 

EXERCISES 

Perform  the  following  indicated  divisions,  and  reduce  the 
resulting  fractions  to  their  lowest  terms : 

gs _p  53    ^    g  +  &  ^    a^-ex-lG  .  a^  +  9a^  +  14 

•    g2_9^>2  •  a-\-3b'  '   x^+4:x-21  '  x^-8x-{-15 

3g4_9a3_54g2     ^  g34-8a^  +  15a 
9a^-117a2  +  378a  *  3a^-33a  +  84*' 


6. 


g-^-llg  +  SO     a^-3a 


8. 


g3_6a2  +  9a      a2-25      a2  +  2a^l5 
a;^-10a;  +  21  .  a.'^-8a;  +  15 

9     a^-6\      b(a-b)       .       6(a  +  5) 
*      a&^a;       a'  +  2ab  +  b'  '  a'-2ab  +  b^' 

10       «  +  ^v       <^'-^'        .    (g-5)2(g  +  5)^ 


11. 


7ar^-56a;-63  *  14a:2^i4^_  1260' 


12     8y^(y  +  4Xy  +  5)  .    y(.v  +  4)(y4-8) 
2X2/  +  5)(2/-7)    •2Xy-7)(2/  +  ll) 

^3     g^(a;-2y     (a;  +  2)(a;-3)  .  a;^ (a;  -  3Xa;  -  5) 
(a; +2)2       (a;-2)(a;-7)  *     (a;-5)(a;-7) 

^4     a'l^ic  +  5)(c- 4)      (c -  8)(c  +  9)  .  a6(c  +  9)(c  -  1) 
(c_4)(c-8)         (c  +  4)(c4-7)  '     (c  +  7)(c  +  l) 

21ar^+23a;-20      6a:^-lla;-10  .  7a^  +  17a;-12 
*     100^2-27 a;  +  5        3a:2  4-2a;-6   ^   5ar^  +  9a;-2  * 


COMPLEX   FRACTIONS  233 

COMPLEX  FRACTIONS 

216.    Sometimes  fractions   occur   whose  numerators  or  de- 
nominators, or  both,  contain  fractions. 

1  +  1        JL^H-L- 

and 


,_i—   1 


a  X  —  1      X  +  1 

Such  fractions  are  called  complex  fractions.  A  complex  frac- 
tion is  said  to  be  simplified  when  it  is  reduced  to  an  equal  frac- 
tion whose  numerator  and  denominator  are  in  the  integraj.  form. 

1+1     «+l 
_,     ^  a         a         a  +  1      a  — 1      a-\-l 

iJiX.  1.  qr  = ^  = 1 = qr* 

l_i     ci  —  1         0,  a         a—1     ^ 

a        a  ^ 

This  result  may  also  be  obtained  directly  by  multiplying  both 

a+  1 


terms  of  the  given  fraction  by  a,  finding  at  once 


a-  1        V 


Ex.2. 


1  1  X  -l-{-x-{-l 

X  -\-  1      X  —  1  ar  —  1 

~I  ~  ~  a;  -f  1  — a-f-  1 


a;-f  1  a^ 


2-     x^^i^=a.. 


x^-l  2 

By  multiplying  the  terms  of  the  given  fraction  by  (a:  -\-  l)(x  —  1) 

we  may  also  get  directly  ^^^ — +  ^  +     _  ^^ 
a:  +  1  —  a:  +  1 

Make  a  rule  for  finding  the  expression  by  which  numerator 
and  denominator  of  a  complex  fraction  may  be  multiplied  so  as 
to  reduce  it  directly  to  a  simple  fraction.  Apply  this  rule  to 
each  of  the  following. 

Ex.  1.  i  +  j-i.     . 

Ex.2.  i±iiLi.  ^       • 


234  ALGEBRAIC   FRACTIONS 


EXERCISES 

Eeduce   each   of    the   following   complex    fractions   to   its 
simplest  form: 

J    l-\-x  m  +  n  +  l  a^-Sb^ 

'l  +  i*  4.  3  7.        27 


X  m  —  n  ~1  3a  — 2b 

2 


2  4 

3.  f.  6. 


a;  g  +  y 

"-2  ~2~ 


a;  — 3      a;  +  3 

•   x±2_x-~2' 

x-3      x+S 


1      +^ 


X   —   4: 
11.        


1 


12.    1^      1 


9. 

c       cD 

1-ht 

c 

a 

13. 

a 

^  +  a  + 

■x 

x-S 

14. 

a;  4-4 
ic  +  2  * 
a;-4 

X-4: 

x  +  2 

15. 

2 
a;-l 

1 

x-2 

3 

4 

1+x  x—S     x—1 


2 

,  2x^  +  2 

x-S 

^-2^ 

1 

1 

16. 

X-S     l-2a;2 


DRILL  EXERCISES 


235 


DRILL  EXERCISES 


Solve  the  following  systems  of  equations  : 
a^  +  2/'  =  25. 


2.   . 

[x-\-2y  =  4. 

•  U--2i/  =  3. 


8. 


11. 


'3x-7by  =  2, 
.2  ax -{- 2  y  =  4:. 


6aj  +  32/  =  l, 
,5ax  —  2by  =  c. 

2x  —  y  —  z  =  Sf 

—  x  —  Sy  +  5z  =  16. 


2»  +  2vH-32  =  10, 
L  —  a;4-2/  +  6z  =  8. 


13.  A  picture  inside  the  frame  is  12  inches  long  and  8 
inches  wide.  If  the  frame  is  a  inches  wide,  express  its  area  in 
terms  of  a. 

14.  If  h  is  the  length  of  the  hypotenuse  of  a  right  triangle 
and  a  the  length  of  one  side,  express  the  length  of  the  third 
side  in  terms  of  h  and  a. 

15.  A  rectangular  piece  of  tin  is  w  inches  wide  and  I  inches 
long.  If  a  square  a  inches  on  a  side  is  cut  out  of  each  corner, 
express  in  terms  of  w,  I,  and  a  the  volume  of  a  box  formed  by 
turning  up  the  sides. 

16.*  A  farmer  plows  a  strip  a  rods  wide  around  a  rectangular 
field  w  rods  wide  and  I  rods  long.  Express  in  terms  of  w,  I, 
and  a  the  area  plowed. 


CHAPTER   XIV 
RATIO,  VARIATION,  AND  PROPORTION 

217.  Definitions.     A  fraction  is  often  called  a  ratio.   Thus  - 

0 

may  be  read  the  ratio  of  a  to  b,  and  is  also  written  a:b. 

The  numerator  is  called  the  antecedent  of  the  ratio,  and  the 
denominator  the  consequent.  The  antecedent  and  consequent 
are  called  the  terms  of  the  ratio. 

An  equation,  each  of  whose  members  is  a  ratio,  is  called 
a  proportion. 

Thus,  -  =  -  is  a  proportion,  and  is  also  written  a:b  =  c:d. 
b      d 

It  is  read  the  ratio  of  a  to  b  equals  the  ratio  ofc  to  d,  or  briefly, 
a  is  to  b  as  c  is  to  d. 

The  four  numbers  a,  6,  c,  and  d  are  said  to  be  in  proportion, 
a  and  d  are  called  the  extremes  of  the  proportion,  and  b  and  c 
the  means. 

218.  If  in  the  graph  of  ?/  =  2fl;,  or-  =  2,  we  think  of  a  point 
as  moving  along  the  line,  we  see  that  its  coordinates  x  and  y 
vary,  but  always  so  that  their  ratio  is  constant,  namely  2. 

Thus  if  xi  and  x^  are  any  two  values  of  x,  and  yi  and  yi  are  the 
corresponding  values  of  y,  given  by  the  equation  y  =  2x,  we  have 
^  =  2  and  ^  =  2,  and  hence  the  proportion  ^  =  ^. 

X\  Xi  X\         Xi 

219.  Definition.  When  any  two  variables  are  connected  in 
such  a  way  that  their  ratio  is  constant,  either  one  is  said  to 
vary  directly  as  the  other.     This  is  usually  written  in  the-form 

where  fe  is  a  constant  and  it  is  read  y  varies  directly  as  x. 

236 


RATIO,   VARIATION,   AND  PROPORTION  237 

In  this  case  the  variables  x  and  y  are  said  tolae  propoHional. 
We  also  say  that  y  is  proportional  to  a;  or  aj  is  proportional 
to  y. 

Ex.  1.  The  formula  for  a  moving  body  s  =  vt  states  that  if  the 
velocity  is  constant,  the  distance  varies  as  the  time.  Hence  if 
^1  and  tz  are  two  values  of  t,  and  s^  and  Sg  the  corresponding 
values  of  s,  we  have  the  proportion 

£l  __  f2,  N^ 

In  this  case  the  space  is  said  to  be  proportional  to  the  time. 

Ex.  2.  From  the  formula  for  percentage,  p  =  br  form  a  pro- 
portion, if  r  is  fixed  and  p  and  b  vary ;  also  if  b  is  fixed  and  p 
and  r  vary. 

Here  the  percentage  is  proportional  to  the  base  if  the  rate 
is  constant,  and  proportional  to  the  rate  if  the  base  is 
constant. 

Ex.  3.  Frojn  the  formula  for  interest,  i—prt  form  a  pro- 
portion if  p  and  r  are  fixed  and  i  and  t  vary ;  if  p  and  t 
are  fixed  and  i  and  r  vary ;  if  r  and  t  are  fixed  and  i  and  p 
vary. 

To  what  is  i  proportional  if  p  and  r  are  fixed  ?  if  r  and  t 
are  fixed  ?   if  p  and  t  are  fixed  ? 

EXERCISES 

1.  If  y  varies  as  x  and  y  =  3  when  x  =  l,  find  y  when  a;  =  12. 

Solution.  We  have  the  proportion  ?(i  ~  ^  in  which  are  given 
x^  =  7,  y^  =  3,  X2  =  12,  to  find  y,-  ^        * 

Substituting,  we  have  -  =  ^  or  yg  =  ^^- 

2.  If  s  varies  as  t  and  if  s  =  40  when  t  =  6,  find  s  when 
t  =  16. 

3.  If  percentage  varies  as  the  base  and  j9  =  50  when  b  =  1000, 
find  b  when  p  =  175. 


238  EATIO,   VAKIATION,   AND  PROPORTION 

IMPORTANT   PROPERTIES   OF  A   PROPORTION 

220.   The  following  examples  show  some  of  the  properties  of 
a  proportion. 

1.  If,  in  the  proportion  -  =  -,  both  members  of  the  equation 
be  multiplied  by  bdj  we  have  ad  =  be. 

That  is  :  If  four  numbers  are  in  proportion,  the  product  of  the 
means  equals  the  product  of  the  extremes. 

2.  Showthat  if-  =  -,  then-  =  -. 

b     d  a      c 

Hint.   Divide  1  =  1  by  the  members  of  the  given  equation. 
This  process  is  called  taking  the  proportion  by  inversion. 


3.    Show  that  if  -  =  -,  then  ^  =  ^ . 
b     d ,  c      d 

Hint.  Multiply  both  members  of  the  given  equation  by  -  • 

c 

To  find  this  multiplier  we  inquire  by  what  expression  - 

a  ^ 

must  be  multiplied  to  give  -  • 

c 

This  process  is  called  taking  the  proportion  by  alternation. 


4.  Show  that  if  «  =  -^  then:5^±*  =  ^±^ . 

b     d  b  d 

Hint.   Add  1  to  both  members  of  the  given  equation. 
This  process  is  called  taking  the  proportion  by  addition. 

5.  Show  that  if  ^  =  -^,  then.^5^  =  ^^  • 

b     d'  b  d 

Hint.   Subtract  1  from  each  member  of  the  given  equation. 
This  process  is  called  taking  the  proportion  by  subtraction. 


IMPORTANT  PROPERTIES  OF  A  PROPORTION         239 

6.  Show  that  if  »  =  f ,  then  ^^+*  =  5+^ . 

b     d  a—b      c—d 

Hint.  Divide  the  members  of  the  equation  obtained  under  4  by 
the  members  of  the  one  obtained  under  5. 

This  process  is  called  taking  a  proportion  by  addition  and 
subtraction. 

7.  Show  that  if  ^  =  :^  =  ^  then  ^L±^±-'  =  «. 

b     d     f  b-]-d^f     b 

Let  -  =  -  =  -  =  A; ;  then  a  =  bk,  c  =  dk.  e  =  fk. 
b     d     f        '  '  '        -^ 

Hence,  a-{-c  +  e  =  bk-\-dk -{-fk  =(b-{-d  +/) k, 

and  ^  +  ^  +  Cfe=^  =  £  =  g. 

6+d+/  b     d     f 

That  is,  If  several  ratios  are  equal,  the  sum  of  the  antecedents 
is  to  the  sum  of  the  consequents  as  any  antecedent  is  to  its 
consequent. 

EXERCISES 

1.    If  ac?  =  be,  show  that  -  =  -.     Hint.  Divide  by  bd. 
b      d 

2    If  ad=6c,  show  that- =  -.     3.  If  ad=&c,show  that  - -r=  -  • 
c     d  c      a 

4.   If  ad  =  be,  show  that  -  =  ^ 


b     a 

.  show  that  — 
b     d 


If  ^  =  -^  show  that  «  +  ^-"  +  ^ 


6. 

"M' 

show  that 

7. 

'n=l' 

show  that 

8. 

b     d 

show  that 

9. 

If?  =  -, 

show  that 

a 

c 

a-b 

c-d 

a 

c 

a-b 

c-d 

a-\-b 

c-^-d 

a^-b 

a-b 

c-\-d 

c-d 

a-\-b  _ 

a 

b     d  c-hd 


240  RATIO,   VARIATION,   AND  PROPORTION 

10.  If2=«,showthat«±^  =  «. 

b     d  b+d     b 

11.  li^  =  2,showth^t^^  =  ^. 

b     d  c—dc 

12.  If  ^  =  i,  show  that  '^^^=^  =  -  • 

b     d'  b-d      b 

13.  Solve  the  equation— =  - for  each  letter  in  terms  of  all 

b     d 
the   others.     If   a  =  3,    6  =  5,  c  =  8,    find   d.     If   6  =  7,   c  =  9, 
d  =  3,  find  a.     If  c  =  13,  d  =  2,  a=^5,  find  b.     If  d  =  50,  a  =  3, 
6  =  -  7,  find  c. 

14.  If  -  =-,  then  x  is  said  to  be  a  fourth  proportional  to  a,  b, 
,  b     X 

and  c. 

Find  a  fourth  proportional  tb  3,  5,  and  7 ;  also  to  9, 5,  and  1, 
and  to  3,  —  2,  and  —  5.' 

15.  If  -  =  -,  then  x  is  called  a  mean  proportional  between 
a  and  o. 

Solve  the  equation  -  =  -  for  x  in  terms  of  a  and  &.     Show 
X      b 
that  there  are  two  solutions,  each  of  which  is  a  mean  propor- 
tional between  a  and  b. 

Find  two  mean  proportionals  between  4  and  9 ;  also  between 
5  and  125,  and  between  —  4  and  —  36. 

,^    TTTi  -T-.T  .         ,.     54-3c?       5  +  4  (?„ 

16.  Which  is  the  greater  ratio,  ^  '       ,  or  ^  '        ? 

^  5+4d       5+5d 

/7^n^     Reduce  the  fractions  to  a  common  denominator  and  com- 
pare numerators,     {d  is  a  positive  number.) 

17.  Which  is  the  greater  ratio, — -  or   — ^— tttt  • 

^  a  +  Sb         aH- 106 

18.  Which  is  the  greater  ratio,  -  or  ^LzL^^  if  b  and  c  are  posi- 

b        b-\-c 

tive,  and  a  less  than  6  ?  a  equal  to  6  ?  a  greater  than  b  ? 


SIMILAR  TRIANGLES  241 

19.  Find  two  numbers  in  the  ratio  of  3  to  5  whose  sum 
is  160. 

20.  Find  two  numbers  in  the  ratio  of  2  to  7  whose  sum  is 
-108. 

21.  Find  two  numbers  in  the  ratio  of  3  to  —  4  whose  sum 
is  - 15. 

22.  What  number  added  to  each  of  the  terms  of  the  ratio 
^  makes  it  equal  to  f^|  ? 

23.  What  number  must  be  added  to  each  term  of  the  ratio 
■^  to  make  it  equal  to  the  ratio  f  ? 

24.  What  number  added  to  each  of  the  numbers  3,  6,  7,  10, 
will  make  the  sums  in  proportion,  when  taken  in  the  given 
order  ? 

25.  Two  numbers  are  in  the  ratio  of  2  to  3,  and  the  sum  of 
their  squares  is  325.     Find  the  numbers. 

SIMILAR   TRIANGLES 

221.   Triangles  are  called  similar  if  they  have  the  same  shape. 

Thus  the  triangles  ABC  and  DBF  of  the  figure  are  similar. 
Note  that  AB  and  DB  have  been  divided  into  7  o 

and  3  equal  parts,  respectively.    Hence  the  ratio 
of  these  sides  is  ^. 

What  is  the  ratio  of  the  sides  BG 
and  BE?    Of  the  sides  CA  and  ED? 
This  is  stated  as  follows  :  The 
lengths  of   the  pairs  of  corre- 
sponding sides  of  two  similar 
triangles  form  a  proportion. 

That  is,  we  may  write  ^  =  ^  =  ^ . 
•^  DB     EB     ED 

Note  that  AB,  BG,  •••  represent  the  lengths  of  these  sides. 


242  RATIO,    VARIATION,   AND  PROPORTION 

EXERCISES 

1.  If  ill  two  similar  triangles  the  sides  in  one  are  11, 13,  and 
16,  and  that  side  in  the  other  which  corresponds  to  the  one 
whose  length  is  11  is  33,  find  the  other  sides  of  the  second 
triangle. 

Solution.     Let  x  represent  the  length  of  the  side  corresponding  to 

the  one  whose  length  is  13.     Then  —  =  — ,  or  a;  =  39. 
^  13      11 

In  this  manner  find  the  remaining  side. 

2.  If  the  sides  of  a  triangle  are  4,  6,  and  10,  and  one  side  of 
a  similar  triangle  is  9,  find  the  remaining  sides  of  the  second 
triangle,  if  the  given  side  corresponds  to  the  side  4. 

3.  Solve  Ex.  2  if  the  given  side  in  the  second  triangle  corre- 
sponds to  6. 

4.  Count   the   number   of   small    triangles   within  each  of 

^  the  triangles  ABC  and  DBE. 

Does  the  result  illustrate  the 
following  statement: 

The  areas  of  similar  trian- 
gles are  in  the  same  ratio  as 
the  squares  of  the  lengths  of 
their  corresponding  sides. 

5.   A   triangular   field    has 

one  side  15  rods  long.     Find 

the  length  of  the  corresponding  side  of  a  similar  field  whose 

area  is  9  times  as  great. 

x^        9 
Suggestion.    The  equation  is  =  -. 

6.  A  triangular  field  has  the  sides  15,  18,  and.  27  rods, 
respectively.  Find  the  dimensions  of  a  similar  field  having 
4  times  the  area. 

7.  The  areas  of  two  similar  triangles  are  49  and  64,  respec- 
tively. One  side  of  the  first  is  12.  Find  the  corresponding 
side  of  the  second. 


SIMILAR  TRIANGLES 


243 


8.  A  triangular  field,  one  of  whose  sides  is  20  rods,  has  an 
area  of  80  square  rods.  Find  the  area  of  a  similar  field  whose 
corresponding  side  is  45  rods. 

It  is  found  in  geometry  that  if  a  line  divides  one  angle  of  a  tri- 
angle into  two  equal  angles,  it  divides  the  opposite  side  into  two 
parts  which  are  in  the  same  ratio  as  the  other  C 

two  sides  of  the  triangle. 

That  is,  in  the  figure  —  =  — . 

9.  Knowing  this  fact,  how  many  of  the 
lines  AD,  DC,  BC,  and  AB  must  you  meas- 
ure in  order  to  find  the  rest  of  them  ?  _b  a 

10.  If  in  the  figure  AD  =  S,  DC  =  6,  and  BC  =  12,  find  AB. 

11.  If  in  the  figure  BC  =  14,  AB  =  18,  and  AD  =  6,  find  DC. 

12.   Ask  and  answer  two  more  questions  like  those 
Exs.  9  and  10. 

This  fact  about  geometry  enables  us  in  some 

cases  to  find  the  distance  between  two  points 

without  measuring  it  directly. 

13.  If  in  the  figure  CD  divides  the 
angle  at  C  into  two  equal  parts  and  if  you 

C        1  A  know  the  lengths  of  AD,  DB,  and  BC, 

show  fully  how  to  find  the  length  of 

a  straight  line  across  the  pond  from  C  to  A  without  measuring 

it  directly. 

14.  Suppose  it  is  found  that  BC=  75  rods,  BD  — 50  rods 
and  AD  =25  rods.     Compute  AC. 

15.  Two  corresponding  sides  of  two  similar  triangles  are  in 
the  ratio  13 :  14.  Show  that  the  perimeters  (sum  of  the  sides) 
of  the  triangles  are  in  the  same  ratio. 

16.  The  perimeters  of  two  similar  triangles  are  in  the  ratio 
33 :  35.  Two  sides  of  the  first  triangle  are  8  and  12.  Find 
two  sides  of  the  second  triangle  corresponding  to  the  given 
sides  of  the  first. 


244  RATIO,   VARIATION,   AND  PROPORTION 

DRILL  EXERCISES 

Simplify : 
1    a?—  1      a^—  3  ■       X 


x-2     4-a^     2-i-x 
2.    ,         }.,         ,  +  1 


(a—b){b  —  c)      {c  —  b)(c  —  a)      {a  —  c)(b  —  a) 
Sab  +  6b^\r  b'-j-a'-^2ab 


3.      a  +  26- 


V  «  +  : 


a -I- 6     J\a^-3ab-^by    a  +  2b 

1^  ^a-^y  •  \2a'  +  3ab  +  b'\ 

a-\-b     a  —  b  x_—_y^  __x-\-jy 

^       a—b      a-{-b  q^    _x y 

a^  -J-  b^     a^  —  b^  x  —  y.  x-\-y' 

(X2  _   52         ^2  _^  ^2  y  3, 


7. 


2         1  1      ]    ^    {a  +  X     a  —  x 


X     a-\-x     a  —  x\        la  — a;     a -\- x 


9.   What    is    the   value    of    ay^-Sx  if  a;  =  l4-V5?     If 

10.   What  is  the  value  of3a^-5fl;4-6ifa;=       ^    ?      If 

x  =  ^-±^? 


2 

W 

a—  Vft, 


11.   What   is    the  value  of  5x'  +  7x  if   a;  =  ^'^'^?     If 


^=       2 

Solve  and  check  by  the  method  of  §  198 : 

12.  a.-2-18a;  +  4  =  0.  15.    2a;2-7a;  =  5. 

13.  x'-Sax  +  b^O,  16.    Sax^-7bx  =  S. 

14.  x^ -]- 9  bx -\- c  =  0.  17.    4  a^^a  —  2  6aa;  =  ca. 

18.  The  edge  of  a  cube  is  a  inches  long.  Express  in  terms  of 
a  (1)  its  volume,  (2)  the  total  area  of  surface,  (3)  the  length  of  di- 
agonal of  one  face,  and  (4)  the  length  of  a  diagonal  of  the  cube. 


CHAPTER   XV 
EQUATIONS  INVOLVING  FRACTIONS 
EQUATIONS  INVOLVING  ALGEBRAIC  FRACTIONS 

222.  We  have  already  seen,  §  101,  that  in  solving  an  equa- 
tion involving  fractions  the  first  step  is  to  multiply  both 
members  of  the  equation  by  a  number  which  will  cancel  all  the 
denominators.  Evidently,  any  common  multiple  of  the  de- 
nominators is  such  a  multiplier.  For  the  sake  of  simplicity, 
and  for  reasons  which  are  fully  discussed  in  the  Adva*nced 
Course,  the  lowest  common  mtUtiple  is  always  used  for  this 
purpose. 

223.  Illustrative  Example.     Solve  the  following  equation : 

2  a;  —  1      __4 3  a;    _  ^  /i  \ 

x-1  ■^a.'-f-l      aJ^-l        '  ^  ^ 

Solution.  The  L.  C.  M.  of  the  denominators  is  x^  —  1.  In  multi- 
plying both  members  by  x^  —  1,  we  see  that  a;  —  1  is  canceled  in  the 
first  fraction,  a:  +  1  in  the  second,  and  a:^  —  1  in  the  third, 

giving          (2a;- l)(r+ 1)  +4(a:-l)-3a:  =  2(ar2-l).  (2) 

Solving,             2a;2  +  x-l  +  4ar-4-3a;  =  2a;2_2.  (3) 

2a:  =  3,  (4) 

and                                                                     X  =  |.  (5) 

Check  by  substituting  ar  =  f  in  (1). 

EXERCISES 

Solve  the  following  equations  and  check  each  solution  by 
substituting  in  the  original  equation,  except  when  the  answer 
is  given : 

'    x  +  1        x-1  '^x'-l         x^-l 
246 


246 


EQUATIONS   INVOLVING  FRACTIONS 
x-1 


Sx  +  5      2a;  +  l^ . 

x-9        x  +  2  "x'-Tx-lS 


A71S.  «  =  f,  or  1. 


3. 


4. 


5. 


30.-4     Ax-l  a^-\-U      ^Q 

aj  +  5        x-j-4:       aj2-f-9a;  +  20 


3aj-15 


2»-10      2a;-6 


3a;2_ii4 


Ax^-S2x-\-60 
6(^+4)  _  3(2.-1)  ^7  _    ^„,^^i^<„_6 

9a^-38 


7. 


8. 


3a;-4  .5a;-7^  

a;_4        2a;-2      2«2-10a;-f 


^Tis.  a;  =  184,  or  2. 


^H-17  _  2(ic+_6) 
a;  +  5         x  +  3 


x-i-S 
x-\-2  ,  3a;-15      3a;-21 


9. 


10. 


11. 


12. 


13. 


14. 


2flj-3     3i»  +  l      4»  +  17 


—  4a;        a;  — 2 


a;-2 


Ans.  X 


19±V395 


3a;-2^2ar^  +  15a;  +  28      2a;-l 
2a;  +  3        2a^+5x  +  S  x  +  1  ' 

2x-'d       x-%  ^  x-\-2 

2x  +  2      5a;  +  2      2a;4-2' 

20a;^  +  7a;-3     3a;4-l^j^ 
9ar*-l  3a;-l 

7a^  +  lla;  +  4       a;  +  3^7a;4-ll 
6ar^  +  13a;  +  5      2x'  +  l       3a;-f5  ' 

3a;4-l       a;-3  ^  2a;^-10a;  +  12 
5aj-7     2a;-7     10 ar* -  49 a;  +  49 


EQUATIONS  IN   ONE   UNKNOWN  24T 

224.  Sometimes  it  is  best  to  add  fractions  before  multiplying 
by  the  L.  C.  M.  and  in  other  cases  to  multiply  by  the  L.  C.  M. 
of  part  of  the  denominators  first,  and,  after  simplifying,  mul- 
tiply by  the  L.  C.  M.  of  the  remaining  denominators. 

Ex.1.     Solve    -i- i-r  =  -^ ^•  (1) 

x  —  2     x-1     x  —  4:     x  —  3 

Adding  fractions  on  the  right  and  left, 

I = 1 (2) 

(x  _  2)(x  - 1)      (x-  4:)(x  -  3)  "^  ^ 

Multiplying  by  L.C.M.,  (x-4:)(x-S)=(x-2)(x-l).     (3) 
Hence,  ^  =  2^.  (4) 

Check  by  substituting  a;  =  2^  in  equation  (1). 

-r^oai                     4^-3^-22^-2  .^. 

Ex.  2.    Solve  —— 7— =  r t:-  (1) 

5  =  f|^2.  (3) 

25^  +  10  =  32^-32.  (4) 

t  =  6.  (5) 

Check  by  substituting  ^  =  6  in  equation  (1). 
In  Exs.  1  and  2,  try  to  splVe  by  clearing  of  fractions  com- 
pletely at  the  outset  and  see  why  the  plan  of  solving  here 
given  is  simpler. 

EXERCISES 

1.    ^^±i-^^i±^  =  ^+L  +  4.  An8.x  =  {i.. 

5  15         6x-S  ^® 

2  '^  a?  + 1      14  a;  -  22  ^  11  g;  +  5 

12  24  8  a; -28* 

3  3^  +  4 _  12^+1  ^5^-1^  ^ns.  x  =  -7f. 

2  8  Sx+2  ^ 

5  15  3^-i-ll  ^ 


Multiplying  by 

16,         4:t- 

-3 

Hence, 

Multiplying  by 

5t  +  2, 

Hence, 

248  EQUATIONS  INVOLVING  FKACTIONS 

lli;-15     33i;  +  15^5'?;  +  5 
10  30  v-5  ' 

gg  —  la;  —  2_a;  —  3      a;  —  4 
x  —  2     3  —  x     x  —  4:     x  —  5' 

1  2^1  ^ 

a;-l     2x4-1      x-2     4x-|-l* 


7. 


a;  — 2  a;  — 3_fl;  — 4  a;  — 5 
a;  — 3  a;  — 4  a;  — 5  6  — a;* 
9  9  5  5 


9. 


x-7     x-2     x-S     a;H-l 


10.    ^  +  '1^  =  ^-^.      ^n..a;=±2V3. 
x  —  2     S  —  x     x—4:     x  —  b 

SIMULTANEOUS  FRACTIONAL  EQUATIONS 
225.   When  pairs  of  fractional   equations   are   given,  each 
should  be  reduced  to  the  integral  form  before  eliminating,  ex- 
cept in  special  cases  like  those  in  the  second  following  illustra- 
tive example. 

Ex.  1.   Solve  the  equations : 

^    +„-l,  =  ::^.  (1) 


x-y     x  +  y     x^-y^ 
3  2  -18 


-.  (2) 

\2x-y     x-3y     {2x-y)(x-Sy) 

From  (1)  by  M,        4(a;  +  y)  +  6(x  -y)  =  36.  (3) 

Byi^,  D,                                       5a;-y  =  18.  (4) 

Prom  (2)  by  M,  3(x -Sy)-2(2x- y)  =  - 18.  (5) 

By  F,D,                                      x  +  7y  =  lS.  (6) 

From  (4)  by  iW,                       S5x-7y  =  126.  (7) 

Adding  (6)  and  (7),                          36  x  =  144.  (8) 

ByD,                                                    a;  =  4.  (9) 

Substitute  a;  =  4  in  (6),                         y  =  2.  (10) 
Check  by  substituting  a;  =  4,  2/  =  2  in  (1)  and  (2). 


SIMULTANEOUS  FRACTIONAL  EQUATIONS 


249 


Ex.  2.     Solve  the  equations : 


[5  +  ^  =  2, 

X      y 

(1) 

2p_21^3 
I  X      y 

(2) 

In  this  case  it  is  best  to  solve  the  equat 

ions 

for 

i 

and 

-^in- 

stead  of  for  x  and  y. 

X 

y 

From  (1)  hj  M,                 ii  +  ?i  =  14. 

X       y 

(3) 

Adding  (2)  and  (3),                   —  =  17. 

X 

(4) 

Hence  by  D,                                -  =  ^' 

X      2i 

(S) 

Substituting  i  =  ^  in  (1),           i  =  |  • 

X    2,                 y    ^ 

(6) 

From  (5)  and  (6)  hj  M,              a?  =  2,  y 

=  3. 

(7) 

In  Ex.  2  try  to  solve  by  first  clearing 

of  fractions  and  see 

why  the  plan  here  used  is  simpler. 

EXERCISES 


Solve  the  following  equations 

2a;-l      3y-l_  ^xy 

x-^1 
1. 


x  +  1 

2/  +  1 

(x  +  l)(y+l/ 

x  +  2 
.2y-l 

2a;-l 
2/4-1  " 

5«y 

(2y- 

-i)(y  +  l) 

3x  +  2 

a^  +  1 

37/-5 

"2/-1' 

3rc-2 

3»-l 

2 

2/  +  1        2/-1       (3/-l)(?/  +  l) 


5     6     „ 

(    V  t 


3,1      o. 

a;     2/ 


5. 


-19. 


1-1  =  12 


6 
12 


2y 


4-^  =  24. 
a      0 


250 


EQUATIONS   INVOLVING  FRACTIONS 


6. 


3_2__ 

-4, 

12_10_^ 

^^  +  ^  =  1, 

X     y 
X      y 

7.    ■ 
52. 

X        y 

H  +  -  =  15. 

U     y 

8.    ■ 

X      2/ 

^  +  ^  =  1. 

[x     y 

X     y 

«    y 

9. 

1  +  1  =  14,           '  10.    < 

y    ^ 

1_l1     ^ 
^  +  -*' 

1  +  1  =  12. 

Z        X 

1  +  1-  = 

Z         X 

c. 

In  Ex.  9  first  add  all  three  equations,  and  from  half  the  sum  sub- 
tract each  equation  separately.     Likewise  in  Ex.  10. 

PROBLEMS  LEADING  TO  FRACTIONAL  EQUATIONS 

In  solving  the  following  problems  use  one  or  two  unknowns 
as  may  be  found  most  convenient. 

1.  There  are  two  numbers  whose  sum  is  51  such  that  if  the 
greater  is  divided  by  their  difference,  the  quotient  is  3^.  Find 
the  numbers.      • 

2.  Find  two  numbers  whose  sum  is  91  such  that  if  the 
greater  is  divided  by  their  difference,  the  quotient  is  7.  Find 
the  numbers. 

3.  There  are  two  numbers  whose  sum  is  s  such  that  if  the 
greater  is  divided  by  their  difference,  the  quotient  is  g.  Find 
an  expression  in  terms  of  s  and  q  representing  each  number. 
Solve  1  and  2  by  substituting  in  the  formula  just  obtained. 

4.  What  number  must  be  subtracted  from  each  term  of  the 
fraction  i^  so  that  the  result  shall  be  equal  to  |-? 

5.  What  number  must  be  subtracted  from  each  term  of  the 
fraction  |^  so  that  the  result  shall  be  equal  to  |-  ? 

6.  What  number  must  be  subtracted  from  each  term  of  the 

fraction  -  so  that  the  result  shall  be  equal  to  -  ?     Solve  4  and 
b  .  ^  d 

5  by  substituting  in  the  formula  obtained  under  6. 


PROBLEMS  LEADING  TO   FRACTIONAL  EQUATIONS      251 

7.  What  number  must  be  added  to  each  term  of  the  frac- 
tion ^  to  obtain  a  fraction  equal  to  \^  ? 

8.  What  number  must  be  added  to  each  term  of  the  fraction 


-  to  obtain  a  fraction  equal  to  -  ? 
0  d 

9.    There  are  two  numbers  whose  difference  is  153.     If  their 
sum  is  divided  by  the  smaller,  the  quotient  is  equal  to  j^. 

10.  There  are  two  numbers  whose  difference  is  cl.  If  their 
sum  is  divided  by  the  smaller,  the  quotient  is  q.  Find  the 
numbers.     Solve  9  by  substituting  in  this  formula. 

11.  Divide  548  into  2  parts,  such  that  7  times  the  first 
shall  exceed  3  times  the  second  by  474.  Ans.  One  part  is 
211.8. 

12.  There  are  two  numbers  whose  sum  is  48  such  that  3 
times  the  first  is  8  more  than  5  times  the  second. 

13.  There  are  two  numbers  whose  sum  is  s  such  that  a  times 
the  first  is  b  more  than  c  times  the  second.     Find  both  numbers. 

14.  What  number  must  be  subtracted  from  each  of  the  num- 
bers 12,  15,  19,  and  25  in  order  that  the  remainders  may  form  a 
proportion  when  taken  in  the  order  given  ? 

15.  What  number  must  be  added  to  each  of  the  numbers  13, 
21,  3,  and  8  so  that  the  sums  shall  be  in  proportion  when  taken 
in  the  order  given  ? 

16.  What  number  must  be  added  to  each  of  the  numbers  a, 
6,  c,  d  so  that  the  sums  shall  be  in  proportion  when  taken  in 
the  given  order? 

17.  What  number  must  be  subtracted  from  each  of  the 
numbers  a,  b,  c,  d  so  that  the  remainders  shall  be  in  proportion 
when  taken  in  the  given  order  ? 

Compare  the  results  in  16  and  17  and  explain  the  relation  between 
them. 

Solve  Exs.  14  and  15  by  substituting  in  the  formulas  obtained  in 
16  and  17. 


252  EQUATIONS  INVOLVING  FRACTIONS 

18.  There  is  a  number  composed  of  two  digits  whose  sum  is 
11.  If  the  number  is  divided  by  the  difference  between  the 
digits,  the  quotient  is  16f.  Find  the  number,  the  tens'  digit 
being  the  larger. 

19.  There  is  a  number  composed  of  two  digits  whose  sum  is 
s.  If  the  number  is  divided  by  the  diiference  between  the 
digits,  the  quotient  is  q.  Find  the  number,  the  tens'  digit 
being  the  larger. 

20.  Illustrative  Problem.  A  can  do  a  piece  of  work  in  8  days, 
B  can  do  it  in  10  days.  In  how  many  days  can  they  do  it 
working  together  ? 

Since  A  can  do  the  work  in  8  days,  in  1  day  he  can  do  I  of  it, 
and  since  B  can  do  it  in  10  days,  in  1  day  he  can  do  -^^  of  it.  If  x 
is  the  number  of  days  required  when  both  work  together,  in  1  day 
they  can  do   -  of  it.     Hence  we  have  the  equation, 

X 

8      10      X 

21.  A  can  do  a  piece  of  work  in  12  days  and  B  can  da  it  in 
9  days.     How  long  will  it  take  both  working  together  to  do  it  ? 

22.  A  pipe  can  fill  a  cistern  in  11  hours  and  another  in  13 
hours.  How  long  will  it  require  both  pipes  to  fill  it  ?  Ans. 
5fJ  hours. 

23.  A  can  do  a  piece  of  work  in  a  days  and  B  can  do  it  in  b 
days.     How  long  will  it  take  both  together  to  do  it  ? 

24.  A  cistern  can  be  filled  by  one  pipe  in  20  minutes  and 
by  another  in  30  minutes.  How  long  will  it 'take  to  fill  the 
cistern  when  both  are  running  together? 

25.  A  pipe  can  fill  a  cistern  in  12  hours,  another  in  10  hours, 
and  a  third  can  empty  it  in  8  hours.  How  long  will  it  require 
to  fill  the  cistern  when  they  are  all  running  ? 

26.  A  man  can  do  a  piece  of  work  in  18  days,  another  in  21 
days,  a  third  in  24  days,  and  a  fourth  in  10  days.  How  long 
will  it  require  them  when  all  are  working  together  ? 

Ans.   4:-^j  days. 


PROBLEMS  LEADING  TO  FRACTIONAL  EQUATIONS   253 

27.  A  and  B  working  together  can  do  a  piece  of  work  in  12 
days.  iB  and  O  working  together  can  do  it  in  13  days,  and  A 
and  C  working  together  can  do  it  in  10  days.  How  long  will 
it  require  each  to  do  it  when  working  alone  ? 

Suggestion :  Let  a  —  the  fraction  of  the  work  A  can  do  in  one  day, 
h  =  the  fraction  of  the  work  B  can  do  in  one  day,  and  c  =  the  fraction 
of  the  work.C  can  do  in  one  day. 

Then,  a  +  6  =  ^,   b  +  c  =  j\,  c  -\-  a  =  ^. 

28.  A  and  B  working  together  can  do  a  piece  of  work  in  I 
days.  B  and  0  can  do  it  in  in  days  and  C  and  A  can  do  it  in 
n  days.     How  long  will  it  require  each  working  alone  ? 

29.  The  circumference  of  the  rear  wheel  of  a  carriage  is  1.8 
feet  more  than  that  of  the  front  wheel.  In  running  one  mile 
the  front  wheel  makes  48  revolutions  more  than  the  rear  wheel. 
Find  the  circumference  of  each  wheel.  # 

If  a;  is  the  number  of  feet  in  the  circumference  of  the  front  wheel, 
then  -= —  is  the  number  of  revolutions  in  going  one  mile. 

X 

30.  The  circumference  of  the  rear  wheel  of  a  carriage  is  1 
foot  more  than  that  of  the  front  wheel.  In  going  one  mile  the 
two  wheels  together  make  920  revolutions.  Find  the  circum- 
ference of  each. 

31.  The  distance  from  Chicago  to  Minneapolis  is  420  miles. 
By  increasing  the  speed  of  a  certain  train  7  miles  per  hour  the 
running  time  is  decreased  by  2  hours.  Find  the  speed  of  the 
train. 

If  r  is  the  original  rate  of  the  train,  then is  the  running  time. 

r 

32.  The  distance  from  New  York  to  Buffalo  is  442  miles.  By 
decreasing  the  speed  of  a  fast  freight  8  miles  per  hour  the  run- 
ning time  is  increased  4  hours.     Find  the  speed  of  the  freight. 

33.  A  motor  boat  goes  10  miles  per  hour  in  still  water.  In 
10  hours  the  boat  goes  42  miles  up  a  river  and  back  again. 
What  is  rate  of  the  current  ? 


254  EQUATIONS   INVOLVING  FRACTIONS 

DRILL  EXERCISES 

Approximate  the  square  roots  of  each  of  the  following  to 
two  places  of  decimals : 

1.  7.9482.  3.   390.07.  5.   .0048. 

2.  4578.9.  4.   9.176.  6.    .04791. 


Solve  the  following  equations : 

J6x-1 
\.S^  — 2 


7.    V2a^-2  =  a;-3.  ^^     V5a;  +  1 


8.   V2a;  +  7=V«  +  3. 


V2a;-3       ^Sx-2 
11.    ■Vx  +  2  =  3--V3-x. 


9.  _V^ — V^  +  2^  =  o.      12.  -^^±i,  =  vio¥+i9. 

Va;  —  a         2  Vet  V8  ic  + 1 


Simplify  the  following: 

13.   ^-y     ^  +  -v  I  y-4g^ , 

a;  —  22!     a;  +  22     4  2;^  —  ic^ 

14  A      a^  +  2a6  +  6Y         ^^^  11^ 

15  ct^  +  27  .  a'b-3ab-h9b 

a^-S    '      a2-f-2a  +  4 

Va;-3     x  +  Aj     \x-{-l     x  —  Sj 

//Tf   Al  n\l/1    _1_  Al  _1_  />\  ni 


{a  —  b  —  c)(a  +  b  -\- c)  '    -i    ,  o^        b-\-a 

/'a-^Y-fie  2  +  ^-^  +  a; 


a6  b'^       b  —  a 

1 

a;- 2 


19.    -9 77C-0 21. 


fa  +  -')+4  X  +  -1--2 

\        aj  aj  +  2 


CHAPTER   XVI 
GENERAL  REVIEW 

226.  The  purpose  of  this  review  is  to  reconsider  some  of  the 
important  topics  of  the  course  in  order  to  show  how  they  are 
interrelated  and  unified  by  a  few  simple  principles^  and  to  gain  a 
little  deeper  insight  into  the  nature  of  algebraic  processes  and 
their  uses. 

FIRST  EXTENSION  OF  THE  NUMBER  SYSTEM 

227.  The  number  system  of  arithmetic  consists  of  the  integers 

tTh 

including  zero  and  the  fractions  of  the  form  — ,  where  m  and  n 

n 

are  integers  (n  not  zero),  together  with  certain  indicated  roots 
such  as  -^.  In  algebra  we  early  encountered  the  negative 
numbers.  These  compelled  us  to  give  a  distinct  name  to  the 
numbers  of  arithmetic  which  were  then  cnWed  positive  numbers. 
The  principles  according  to  which  positive  and  negative  numbers 
are  added,  subtracted,  multiplied,  and  divided  were  studied  in 
connection  with  concrete  facts  to  which  these  numbers  are 
naturally  applicable. 

EXERCISES 

1.  Do  negative  numbers  apply  to  all  things  to  which  posi- 
tive numbers  apply  ?  Can  there  be  a  negative  number  of 
books  on  a  shelf? 

2.  Make  a  list  of  things  to  which  both  positive  and  negative 
numbers  apply. 

3.  Show  by  concrete  examples  different  from  those  in  this 
book  how  Principle  VII  is  obtained. 

265 


256  GENERAL   REVIEW 

4.  In  arithmetic  how  did  you  find  the  difference  in  temper- 
ature between  8°  above  zero  and  40°  above  zero  ?  Between  8° 
above  zero  and  10°  below  zero  ?  Note  that  in  the  latter  case 
you  added  to  find  a  difference.  How  does  this  compare  with 
the  subtraction  of  a  negative  number  from  a  positive  or  a  posi- 
tive number  from  a  negative  ?  By  concrete  examples  verify 
Principle  VII.  Show  how  a  problem  in  subtraction  may  be 
changed  into  a  problem  in  addition  by  means  of  Principle  VIII. 

5.  Verify  Principle  IX  by  concrete  examples. 

6.  How  may  the  signs  of  the  factors  of  a  product  be 
changed  without  changing  the  value  of  the  product  ?  Make 
all  possible  changes  of  sign  in  {a  —  b)(b  —  c)(c  —  d)  which  will 
not  change  its  value.  Also  make  all  possible  changes  of  sign 
which  will  change  the  sign  of  the  product. 

7.  Define  division  in  terms  of  multiplication.  Derive 
Principle  X  from  Principle  IX  by  means  of  the  definition  of 
division. 

8.  State  how  the  signs  involved  in  a  fraction  may  be 
changed  without  changing  the  value  of  the  fraction.  Make 
all  possible  changes  of  signs  which  will  not  change  the  value 

of  -,  also  of  --. 
b  b 

9.  Make  all  possible  changes  of  signs  which  will  leave 

unchanged  the  value  of  the  expression  v^  ~  y>  ^^  —  z){z  —  x)  ^ 

^  (a-b){b-c) 

10.  Eeduce  ^^^,  ^^=^,  and  ^i-^ to  fractions  hav- 

x-y-  y-z  (y-x){z-y) 

ing  a  common  denominator. 

11.  Eeduce , ,  and 

(a  -  b)(b  -cY  (c-  d)(b  -  ay  (c  -  b){d  -  c) 

to  fractions  having  a  common  denominator. 

12.  Recall  some  problem  solved  during  the  year,  in  which 
a  negative  result,  while  satisfying  the  equation,  does  not  have 
meaning  in  the  problem. 


SECOND   EXTENSION  OF   THE  NUMBER   S.YSTEM       257 

228.  Rational  and  Irrational  Numbers.  The  positive  and  neg- 
ative integers  and  fractions,  including  zero,  are  called  rational 
numbers.  In  attempting  to  solve  the  equation  a^  =  2  we  get 
a;=  ±  V2  in  which  V^  stands  for  a  number  whose  square 
is  2.  It  may  be  shown  that  no  rational  number  satisfies  this 
condition.     Such  a  number  is  called  an  irrational  number. 

In  this  book  the  only  irrational  numbers  encountered  are 
irrational  square  roots  called  quadratic  surds,  whose  values  we 
learned  to  approximate  to  any  desired  degree  of  accuracy. 
Other  forms  of  irrational  numbers  are  found  in  more  advanced 
work. 

The  essential  property  of  a  quadratic  surd  is  stated  in 
principle  XVIII. 

SECOND  EXTENSION  OF  THE  NUMBER  SYSTEM 

229.  The  equation  ar^  +  l  =  0ora2  =  — 1  introduces  still  an- 
other kind  of  number,  namely,  x—  ±  V— 1.  Such  numbers, 
called  imaginaries,  were  met  in  the  solution  of  quadratic  equa- 
tions, but  their  study  was  postponed  to  the  Advanced  Course. 

The  irrational  and  imaginary  numbers  form  important  topics 
in  higher  courses  in  mathematics. 

The  extension  of  the  number  system  to  include  negative  num- 
bers and  imaginaries,  besides  the  rational  and  irrational  numbers 
of  arithmetic,  is  one  of  the  chief  distinctions  between  arithmetic 
and  algebra. 

LITERAL  EQUATIONS 

230.  Identities.  There  are  two  essentially  different  kinds 
of  literal  equations.  One  is  the  identity  of  which  an  example 
is  a(h-\-c)  =ab-{-ac. 

If  all  the  indicated  operations  in  an  algebraic  identity  are 
performed,  the  two  members  become  exactly  alike.  Hence  the 
chief  use  of  identities  is  to  formulate  those  operations  on  alge- 
braic expressions  which  change  their  form  hut  not  their  value. 


258  .  GENERAL  REVIEW 

EXERCISES 

1.  What  values  of  the  letters  involved  in  an  identity  will 
satisfy  it? 

2.  State  in  the  form  of  identities  as  many  as  possible  of  the 
eighteen  principles  given  in  this  book. 

Determine  which  of  the  following  are  identities: 

3.  {x^-yY  =  x'-^2xy  +  y\     5.    {x-{-lf  =  x'-2. 

4.  {x-yy  =  a^-2xy  +  y'.    6.    a^ -^  ab  +  b^  =  (a -\- by  -  ab. 

7.  a^-b'-  =  (a-b)(a-\-  b). 

8.  (a  -  b)(a'  +  «6  +  b^)  =a« -  ¥. 

9.  (a+b)(a''-ab-\-b^)=a^-^b\ 

10.  (x-\-  b)  (x  +  a)  =  x^ -]- (a -\-  b)x  +  ab. 

11.  {x  —  a){x  +  b)=:x^—{a-\-b)x-\-ab. 

12.  {a  +  b-cy=za'-\.b''-\-^  ■\-2ab-2ac-2bc. 

231.  Equations  of  Condition.  The  other  kind  of  literal  equa- 
tion is  one  which  is  not  satisfied  for  all  values  of  all  the  letters 
involved.     These  originate  in  a  large  variety  of  ways. 

For  example  in  />  =  hr,  which  may  be  regarded  as  a  definition  of 
percentage,  any  value  may  be  assigned  to  two  of  the  letters  but  the 
third  is  thereby  determined.  Likewise  in  i  =  pr-t,  which  may  be  re- 
garded as  a  definition  of  interest,  any  value  may  be  given  to  three  of 
the  letters  thereby  fixing  the  fourth. 

In  the  equation  w\di  =  w^d^,  which  is  the  law  of  the  lever  and  was 
established  by  experimentation,  any  three  letters  may  be  given  arbi- 
trarily, thereby  fixing  the  fourth. 

If  in  a  literal  equation  numerical  values  are  assigned  to  all 
the  letters  except  one,  then  the  value  of  this  one  is  determined 
by  the  equation. 

232.  In  general,  every  literal  equation  which  is  not  an  iden- 
tity states  a  certain  relation  that  must  exist  between  the  vari- 
ables involved.  In  this  sense  such  an  equation  is  an  equation 
of  condition. 


LITERAL   EQUATIONS  259 

In  using  a  literal  equation  it  may  become  necessary  to  solve 
it  for  any  one  of  the  letters  in  terms  of  the  others,  and  we  thus 
have  an  important  example  of  the  solution  of  equations  in  one 
unknown. 

EXERCISES 

1.  State  as  many  rules  of  arithmetic  as  you  can  in  the 
form  of  literal  equations  and  indicate  which  of  these  are  iden- 
tities and  which  are  equations  of  condition. 

2.  Indicate  which  of  the  equations  of  the  preceding 
example  can  be  derived  from  others. 

3.  Enumerate  the  various  steps  which  may  be  used  in  the 
solution  of  an  equation  in  one  unknown. 

4.  Solve  for  x  and  y  the  simultaneous  equations 

( (hx  +  bjy  =  c^, 
1  ajjO?  +  ^22/  =  Cz, 
and  reduce  the  results  to  the  simplest  form. 

Note.  —  The  letters  a^,  a^  6j,  h^  are  read  a  one,  a  two,  b  one,  b  two, 
etc.  The  numbers  a^  and  a,,  etc.,  are  entirely  distinct  so  far  as  their 
values  are  concerned,  the  subscripts  1  and  2  being  used  to  distinguish 
two  different  coefficients  of  the  same  letter  x. 

In  solving  these  equations  the  steps  involved  are  exactly  the  same 
as  are  used  in  solving  any  other  pair  of  linear  equations.  The  result 
obtained  may,  therefore,  be  used  as  a  formula  from  which  the  solution 
of  any  two  such  equations  may  be  obtained. 

5.  Solve  equations  1  to  5  on  page  119  by  means  of  this 
formula. 

6.  What  equations  would  have  to  be  solved  to  obtain  a 
formula  for  the  solution  of  three  equations  in  three  unknowns? 

7.  What  statements  can  you  make  about  the  relation  of  the 
equation  ax-  +  6a;  +  o  =  0  to  any  other  quadratic  equation? 


260  GENERAL   REVIEW 

VARIABLES,  GRAPHS,  FUNCTIONALITY 
233.   We  have  seen  that  an  equation  like  s  =  vt,  where  v  is  a 
constant,  expresses  a  fixed  relation  between  the  variables  s  and  t. 

Thus,  if  y  =  8,  thens=  St,  aud  the  variables  change  subject  to 
the  functional  relation  expressed  by  this  equation. 

The  study  of  the  dependence  of  one  variable  on  other  vari- 
ables is  to-day  engaging  the  best  effort  of  great  men  in  all  the 
sciences.  The  attempt  is  constantly  made  in  each  case  to  so 
determine  the  nature  of  such  dependence  that  it  may  be  ex- 
pressed in  the  form  of  a  literal  equation. 

The  cases  of  functional  relations  between  variables  which 
have  been  studied  in  this  book  are  but  a  very  few  of  the  sim- 
plest kind  and  are  confined  to  two  variables  only.  The  graph 
is  a  convenient  means  of  exhibiting  a  known  relation  between 
two  variables. 

EXERCISES 

1.  Write  an  equation  of  the  first  degree  in  two  variables  x 
and  y  in  which  y  increases  as  x  increases. 

2.  Construct  a  graph  representing  the  equation  of  the  pre- 
ceding. 

3.  Write  a  similar  equation  in  which  y  decreases  as  x  in- 
creases. 

4.  Construct  a  graph  representing  the  equation  obtained  in 
the  preceding. 

5.  How  many  points  do  you  need  to  locate  on  the  graph  of 
a  linear  equation  before  you  can  draw  the  whole  graph  ? 

6.  Construct  graphs  of  the  equations  3  a;  —  2  2/  =  12  and 
2a;  =  5?/+10. 

7.  What  is  the  relation  between  the  graphs  of  the  equations 
2a;-f-32/  =  6and2a;  +  32/  =  12? 


SIMULTANEOUS  EQUATIONS  261 

SIMULTANEOUS  EQUATIONS 

234.  We  have  seen  that  one  equation  in  two  variables  may 
be  regarded  as  stating  a  relation  between  these  variables  sub- 
ject to  which  they  are  permitted  to  vary. 

Thus  in  the  equation  x  +  y  =  5,x  and  y  may  vary  at  will  so  long  as 
their  sum  remains  5. 

Such  an  equation  is  satisfied  by  an  endless  number  of  pairs 
of  values.  While  this  is  true  of  each  of  two  equations  in  two 
variables,  there  is  only  one  pair  of  values  which  satisfy  both 
equations,  provided  the  equations  are  independent  and  simulta- 
neous and  both  of  the  first  degree.  Hence  two  linear  equations, 
each  in  two  variables,  completely  determine  these  variables. 
This  is  most  easily  seen  by  means  of  the  graph.    See  page  112. 

The  solution  of  a  pair  of  simultaneous  linear  equations  con- 
sists in  finding  this  one  pair  of  numbers  which  satisfy  both 
equations. 

EXERCISES 

1.  By  means  of  an  example  explain  elimination  by  addition 
and  subtraction. 

2.  By  means  of  an  example  explain  elimination  by  substi- 
tution. 

3.  Under  what  conditions  is  elimination  by  substitution 
simpler  than  elimination  by  addition  and  subtraction  ?  Under 
what  conditions  is  it  not  ? 

FACTORING,  QUADRATICS  AND  FRACTIONS 

235.  To  be  able  to  select  at  sight  the  factors  of  simple  algebraic 
expressions  is  a  prerequisite  for  successful  and  effective  algebraic 
manipulation.  It  will  be  well  to  review  the  various  types  of 
factoring  given  in  the  text.  You  should  now  be  able  to  read 
at  sight  the  factors  of  many  of  the  exercises  in  Chapter  IX, 
and  to  solve  at  sight  most  of  the  equations  on  page  168,  after 
reducing  each  to  the  type  form  aoc^  -\-bx  +  c  =  0. 


262  GENERAL  REVIEW 

EXERCISES 

1.  Make  a  list  of  the  various  forms  of  binomials  which  are 
factorable,  and  show  how  to  find  the  factors. 

2.  Give  a  similar  treatment  of  trinomials. 

3.  What  forms  of  polynomials  of  four  terms  are  you  able  to 
factor  ? 

4.  Can  you  factor  the  following  polynomials  of  five  terms : 
(1)  a2  +  2a6  +  &-  +  a  +  6?  (2)  4ar^  + 12 a;?/ +  9/+ 2a; +  3 2/? 

5.  In  the  exercises  on  pages  143-146  and  152-155  find  as 
many  as  you  can  of  the  results  at  sight. 

6.  In  the  exercises  on  pages  228-232  find  as  many  as  you 
can  of  the  results  at  sight.  You  should  at  least  be  able  to 
write  the  results  without  first  writing  in  full  all  the  factors  of 
each  polynomial. 

7.  What  is  the  form  of  the  square  of  a  polynomial.  Ex- 
plain the  method  of  finding  the  square  root  of  a  polynomial 
and  also  of  a  number  expressed  in  arabic  figures. 

236.  Factors  of  Quadratic  Expressions.  We  have  seen  that 
certain  quadratic  equations  may  be  solved  by  means  of  fac- 
toring. 

For  example,  z^  —  7  x  +  12  —  0,  in  which  x  —3  and  x  —  4  are  the 
factors  of  the  left  member  and  3  and  4  are  the  roots. 

It  is  now  possible  to  use  the  general  solution  of  any  quadratic 
equation  a^  -{-px  -\-q  =  0  for  the  purpose  of  factoring  the  tri- 
nomial a^  -{-px  +  q. 

In  a^  —  7  a;  -|- 12  =  0  we  observe  that  4  is  a  root  of  the  equa- 
tion and  that  a;  —  4  is  a  factor  of  the  left  member.  Also  3  is  a 
root  and  a;  —  3  is  a  factor. 

In  like  manner  if  ri  and  rg  are  the  roots  of  the  quadratic 
equation  a:^-\-px-{-  q  =  0,  then  x  —  ri  and  x  —  r2  are  the 
factors    of    a^  -^  px  -\-  q. 


FACTORING,    QUADRATICS,   AND  FRACTIONS  263 

Example  :     Factor  x"^  +  2x—^. 

Solution :  Solviiig  the  equation  x2  +  2x  —  5  =  0  we  get  x  =  —  \ 
+  V6  and  x  =  —  \  —  V6.  Hence  the  factors  of  x^  +  2  a;  —  5  are 
X  -  (-  1  +  V6)  and  x  -  (-  1  -  V6). 

Thatis,  x2+2a:-5:=(a:  +  l-  V6)  (x  +  l  +  \/6). 

Verify  by  multiplying  the  factors  together  to  see  if  the 
product  is  a;^  +  2  a;  —  5.  The  multiplication  of  the  factors  thus 
found  may  be  used  as  a  check  on  the  solution  of  the  quadratic. 

EXERCISES 

Factor  by  the  above  method  the  following  quadratic  ex- 
pressions and  verify  the  results  by  multiplying  the  factors: 

1.  a^  +  7a;-4.  4.   a^-Saj  +  l.         7.   a?  +  ax-h. 

2.  ar^  +  12aj  +  13.         5.   a?-^x-^.         8.   a?-2hx-5h. 

3.  ar^-9a;  +  12.  6.   a?-ax-\-h.  9.  V-l-4aa;  +  2a. 

OPERATIONS   AT   SIGHT 

237.  You  should  now  be  sufficiently  familiar  with  the  simple 
algebraic  operations  to  carry  out  many  of  them  without  writing 
down  each  step.     This  is  illustrated  in  the  following  examples. 

Ex.  1.     Solve45a;-14  +  7rc  =  13x  +  7  +  8x  +  41. 

Solution :  31  x  =  62  and  x  =  2. 

We  add  45  x  and  7x  and  then  subtract  \3x  and  8x  (the  latter  are 
to  be  subtracted  from  both  member^  of  the  equation)  getting  31  x. 
Then  we  add  14,  7,  and  41  (14  must  be  added  to  each  member  of  the 
equation)  getting  62.     All  this  is  done  mentally. 

In  this  manner  solve  Equations  1  to  20,  pages  26,  27. 

Ex.  2.  Add  the  polynomials  3  ax  +  5  by -S  6V,  4  W_2  ax 
-  7  by,  and  4ax  -  2  &2/  -}-  7  6V. 

Solution :  We  pick  out  at  once  all  terms  having  the  factor  ax  and 
add  them,  then  the  terms  having  the  factor  by,  and  finally  the  terms 
having  the  factor  b^c^  obtaining  5  ax  —  4  J?/  +  3  b^c^. 

Polynomials  may  be  subtracted  in  the  same  way. 


264  GENERAL  REVIEW 

In  this  manner  solve  Exs.  1  to  18,  page  61. 

Ex.3.     Solve  jf^  +  ^2/=14,  (1) 

\5x-3y  =  S.  (2) 

Solution  :  25  y  +  9  ?/  =  70  -  24  =  46  and  ?/  =  46  -^  34  =  f  f . 

We  notice  that  to  eliminate  x  we  must  multiply  equation  (1)  by 
5  and  equation  (2)  by  3  and  then  subtract.  Since  this  will  eliminate 
X,  we  need  not  write  down  the  terms  containing  it.  Hence  we  get 
5  X  5y-3(-3?/)  =  5  x  14  -  3  x  8,  or  34y  =  46. 

In  this  manner  solve  Exs.  10  to  20,  page  117. 
The  following  operations  should,  for  the  most  part,  be  done 
at  sight,  or  with  very  little  writing : 

(1)  Finding  the  products  of  any  two  binomials. 

(2)  Practically  all  the  factoring  in  this  book. 

(3)  Supplying  the  third  term  to  complete  any  trinomial 
square,  when  two  terms  are  given. 

(4)  Solving  a  quadratic  by  means  of  the  formula. 

(5)  Most  of  the  cancellations  needed  in  dealing  with  fractions. 

(6)  Selection  of  the  rationalizing  factor  for  a  quadratic  surd 
and  applying  it. 

Turn  now  to  these  various  topics  and  consider  each  exercise 
a  challenge  to  your  mental  skill. 

Such  a  familiarity  with  the  operations  of  algebra  as  is  here 
indicated  will  greatly  aiH  you,  not  only  in  your  future  work  in 
mathematics,  but  also  in  the  study  of  a  science  such  as  physics. 

PROBLEMS 

238.  The  solution  of  problems  by  means  of  equations  is  an 
important  feature  of  algebra.  It  is  a  much  more  powerful 
method  than  arithmetic  provides.  Even  though  the  operations 
be  actually  the  same  in  both  cases,  yet  the  equation  enables  us 
to  tabulate  the  data  of  a  problem  in  such  a  way  that  they 
can  be  handled  clearly  and  concisely  and  so  that  we  may  ob- 
tain the  solution  by  means  of  a  few  standard  operations. 


PROBLEMS  265 

239.  Problems  have  in  all  ages  held  a  peculiar  interest  for 
the  human  mind.  Some  are  of  interest  because  of  the  chal- 
lenge they  present  to  the  mind  as  mere  puzzles.  Such  is  the 
following : 

Ex.  1.  A  father  is  9  years  less  than  twice  as  old  as  his  son  and  12 
years  ago  he  was  19  years  less  than  three  times  as  old  as  the  son. 
Find  the  age  of  each. 

Other  problems  stated  in  the  puzzle  form  attract  attention 
because  of  the  interesting  facts  revealed  by  their  solutions. 
For  example : 

Ex.  2.   The  Fahrenheit  reading  at  the  temperature  of  liquid  air  is 

128  degrees  lower  than  the  Centigrade  reading.     Find  both  the  Centi- 
grade and  the  Fahrenheit  reading  at  this  temperature. 

Still  other  problems,  while  mere  puzzles,  are  of  value  because 
they  deal  with  important  scientific  or  mathematical  data.  For 
example : 

Ex.  3.  A  boatman  trying  to  row  up  a  river  drifts  back  at  the  rate 
of  1^  miles  per  hour,  while  he  can  row  down  the  river  at  the  rate  of 
12  miles  per  hour.     What  is  the  rate  of  the  current  ? 

Ex.  4.  A  beam  is  12  feet  long.  It  carries  a  40-pound  weight  at 
one  end,  a  60-pound  weight  3  feet  from  this  end,  and  a  70-pound 
weight  at  the  other  end.  Where  is  the  fulcrum  if  the  beam  is 
balanced  ? 

Ex.  5.  There  is  a  rectangle  whose  length  is  60  feet  more,  and  whose 
width  is  20  feet  less,  than  the  side  of  a  square  of  equal  area.  Find 
the  dimensions  of  the  square  and  the  rectangle. 

But  by  far  the  most  important  problems  are  those  which  it 
is  necessary  for  mankind  to  solve  in  order  to  gain  certain  de- 
sired information.  Such  problems  are  called  real  problems  or 
practical  problems.     For  example  : 

Ex.  6.  A  farmer  has  a  field  120  rods  long  and  60  rods  wide.  How 
wide  a  strip  must  he  plow  around  it  to  make  5  acres? 


266  GENERAL  REVIEW 

Ex.  7.  Find  the  aveiage  temperature  for  12  hours  from  the  fol- 
lowing hourly  observations  ;  —  10^  ;  -  7°  ;  —  5° ;  —  5°  ;  —  1° ;  +  3° ; 
^70;   +9<=";   +90.   ^40.  _^2°;   -4°. 

Ex.  8.  The  earth  and  Mars  were  in  conjunction.  When  are  they 
next  in  conjunction  if  the  earth's  period  is  365  days  and  that  of  Mars 
687  days  ?     (See  figure,  page  269.) 

240.  The  development  of  every  physical  science  depends  to 
a  large  extent  upon  the  solution  of  problems  of  this  latter  type, 
and  algebra  is  one  of  the  great  tools  for  this  purpose.  It  is 
not  possible  for  you  at  this  stage  to  apply  algebra  to  the  solution 
of  many  kinds  of  problems  of  this  type  because  of  your  limited 
knowledge  of  the  scientific  principles  involved  in  their  state- 
ment. But  if  you  have  mastered  the  various  processes  of 
solving  equations,  if  you  are  familiar  vi^ith  factoring  and 
fractions,  with  square  root  and  radicals,  in  short,  with  all  the 
processes  used  in  this  course,  you  are  thus  in  possession  of  the 
tools  with  which  to  solve  any  problems  whose  conditions  you 
can  translate  into  statements  in  the  form  of  linear  or  quadratic 
equations. 

SUPPLEMENTARY  PROBLEMS  ON  CHAPTER  II 

1.  The  area  of  Louisiana  is  (nearly)  4  times  that  of  Mary- 
land, and  the  sum  of  their  areas  is  60,930  square  miles.  Find 
the  (approximate)  area  of  each  state. 

2.  The  horse  power  of  a  certain  steam  yacht  is  12  times 
that  of  a  motor  boat.  The  sum  of  their  horse-powers  is  195. 
Find  the  horse  power  of  each. 

3.  At  a  football  game  there  were  2000  persons.  The  num- 
ber of  women  was  3  times  the  number  of  children,  and  the 
number  of  men  was  6  times  the  number  of  children.  How 
many  men,  women,  and  children  were  there  ? 

4.  It  is  twice  as  far  from  New  York  to  Syracuse  as  from 
New  York  to  Albany,  and  it  is  4  times  as  far  from  New 
York  to  Cleveland  as  from  New  York  to  Albany.  The  sum  of 
the  three  distances  is  1015  miles.     Find  each  distance. 


SUPPLEMENTARY  PROBLEMS  ON  CHAPTER  II   267 

5.  The  altitude  of  Popocatepetl  is  1716  feet  less  than  that 
of  Mt.  Logan,  and  the  altitude  of  Mt.  St.  Elias  is  316  feet 
greater  than  that  of  Popocatepetl.  Find  the  altitude  of  each 
mountain,  the  sum  of  their  altitudes  being  55,384  feet. 

6.  It  is  4  times  as  far  from  New  York  City  to  Cincinnati 
as  from  New  York  to  Baltimore.  Twice  the  distance  from 
New  York  to  Cincinnati  minus  5  times  that  from  New  York 
to  Baltimore  equals  567  miles.  How  far  is  it  from  New  York 
to  each  of  the  other  cities  ? 

7.  The  melting  temperature  of  glass  is  276  degrees  (Centi- 
grade) higher  than  twice  that  of  zinc.  One-half  the  number  of 
degrees  at  which  glass  melts  plus  7  times  the  number  at  which 
zinc  melts  equals  3434.      Find  the  melting  point  of  each. 

8.  The  melting  temperature  of  nickel  is  496  degrees  (Centi- 
grade) higher  than  that  of  silver.  Three  times  the  number  of 
degrees  at  which  nickel  melts  plus  2  times  the  number  at  which 
silver  melts  equals  6258.     Find  the  melting  point  of  each. 

9.  A  cubic  foot  of  nickel  weighs  80  pounds  more  than  one 
cubic  foot  of  tin.  Four  cubic  feet  of  nickel  plus  2  cubic  feet 
of  tin  weigh  3056  pounds.  Find  the  weight  per  cubic  foot  of 
each  metal. 

10.  A  cubic  foot  of  gold  weighs  545  pounds  more  than  a 
cubic  foot  of  silver.  One  cubic  foot  of  gold  and  one  cubic  foot 
of  silver  together  weigh  1855  pounds.  Find  the  weight  per 
cubic  foot  of  each  metal. 

11.  A  cubic  foot  of  steel  weighs  17  times  as  much  as  a  cubic 
foot  of  yellow  pine.  The  combined  weight  of  11  cubic  feet  of 
pine  and  3  cubic  feet  of  steel  is  1773.2  pounds.  Find  the 
weight  of  one  cubic  foot  of  each. 

12.  The  area  of  Great  Britain  is  7557  square  miles  more 
than  9  times  that  of  the  Netherlands,  and  the  area  of  Japan  is 
42,065  square  miles  less  than  15  times  that  of  the  Netherlands. 
One-third  the  area  of  Great  Britain  plus  -J-  the  area  of  Japan  is 
69,994  square  miles.     Find  the  area  of  each  country. 


268  GENERAL   REVIEW 

13.  The  diameter  of  the  earth  is  1918  miles  more  than 
twice  that  of  Mercury,  and  the  diameter  of  Venus  is  1700 
miles  more  than  twice  that  of  Mercury.  The  diameter  of  the 
earth  plus  -J-  that  of  Venus  equals  11,768  miles.  Find  the 
diameter  of  each  planet. 

14.  The  diameter  of  Jupiter  is  500  miles  more  than  20 
times  that  of  Mars,  and  the  diameter  of  Saturn  is  4200  miles 
more  than  16  times  that  of  Mars.  One-tenth  the  diameter  of 
Jupiter  plus  ^  that  of  Saturn  is  45,150  miles.  Find  the 
diameter  of  each  planet. 

15.  The  diameter  of  Neptune  is  2900  miles  more  than  that 
of  Uranus.  The  sum  of  their  diameters  is  66,700  miles. 
Find  the  diameter  of  each  planet. 

16.  The  money  circulation  of  the  United  States  in  1880  was 
136  million  dollars  more  than  3  times  that  in  1850.  In  1910 
it  was  51  million  more  than  11  times  that  in  1850.  The 
circulation  of  1910  exceeded  that  of  1880  by  2147  million. 
Find  the  circulation  in  each  year. 

SUPPLEMENTARY  PROBLEMS  ON  CHAPTER  VI 

1.  A  sparrow  flies  135  feet  per  second  and  a  hawk  149  feet 
per  second.  The  hawk  in  pursuing  the  sparrow  passes  a  cer- 
tain point  7  seconds  after  the  sparrow.  In  how  many  seconds 
from  this  time  does  the  hawk  overtake  the  sparrow  ? 

2.  A  courier  starts  from  a  certain  point,  traveling  Vi  miles 
per  hour,  and  a  hours  later  a  second  courier  starts,  going  at 
the  rate  of  Vg  miles  per  hour.  In  how  long  a  time  will  the 
second  overtake  the  first,  supposing  V2  greater  than  Vi  ? 

If  the  second  courier  requires  t  hours  to  overtake  the  first,  the  latter 
had  been  on  the  way  t  +  a  hours.  Thus  the  distance  covered  by  the 
second  courier  is  Vg^  and  by  the  first  Vi(t  +  a).  As  these  numbers  are 
equal  we  have  if,i  =  v^(t -\- a). 

This  formula  summarizes  the  solution  of  all  problems  like  3  and 
4  below. 


SUPPLEMENTARY  PROBLEMS  ON  CHAPTER  VI   269 

3.  In  an  automobile  race  A  drives  his  machine  at  an 
average  rate  of  53  miles  per  hour,  while  B,  who  starts  \  hour 
later,  averages  57  miles  per  hour.  How  long  does  it  require  B 
to  overtake  A  ?     Use  the  above  formula. 

4.  A  freight  steamer  leaves  New  York  for  Liverpool,  aver- 
aging 10|-  knots  per  hour,  and  is  followed  4  days  later  by  an 
ocean  greyhound,  averaging  25^  knots  per  hour.  In  how  long 
a  time  will  the  latter  overtake  the  former  ? 

5.  One  athlete  makes  a  lap  on  an  oval  track  in  26  seconds, 
another  in  28  seconds.  If  they  start  together  in  the  same 
direction,  in  how  many  seconds  will  the  first  gain  one  lap  on 
the  other?     Two  laps? 

Let  one  lap  be  the  unit  of  distance.  Since  the  first  covers  one  lap 
in  26  seconds,  his  rate  per  second  is  ^,  Likewise  the  rate  of  the  other 
is  ^j.  If  t  is  the  required  number  of  seconds,  the  distance  covered  by 
the  first  is  2^  t  and  by  the  second  ^  t.  If  the  first  goes  one  lap  farther 
than  the  second,  the  equation  is  ^  <  =  ^^  <  4  1. 

6.  Two  automobiles  are  racing  on  a  circular  track.  One 
makes  the  circuit  in  31  minutes  and  the  other  in  38 J  minutes. 
In  what  time  will  the  faster  machine  gain  1  lap  on  the  slower? 

7.  The  planet  Mercury  makes  a  cir- 
cuit around  the  sun  in  3  months  and 
Venus  in  7^  months.  Starting  in  con- 
, Venus  junction,  as  in  the  figure,  how  long  be- 
fore they  will  again  be  in  this  position  ? 
Note  that  the  problem  may  be  solved  the 
same  as  if  the  two  planets  were  moving  in 
the  same  orbit  at  different  rates. 

8.  Saturn  goes  around  the  sun  in  29  years  and  Jupiter  in 
12  years.  Starting  in  conjunction,  how  soon  will  they  be  in 
conjunction  again  ? 

9.  Uranus  makes  the  circuit  of  its  orbit  in  84  years  and 
Neptune  in  164  years.  If  they  start  in  conjunction,  how  long 
before  they  will  be  in  conjunction  again  ?    Ans,  172^  years. 


270  GENERAL  REVIEW 

10.  The  hour  hand  of  a  watch  makes  one  revolution  in  12 
hours  and  the  minute  hand  in  1  hour.  How  long  is  it  from 
the  time  when  the  hands  are  together  until  they  are  again 
together  ? 

11.  One  object  makes  a  complete  circuit  in  a  units  of  time 
and  another  in  h  units  (of  the  same  kind).  In  how  many  units 
of  time  will  one  overtake  the  other,  supposing  h  to  be  greater 
than  a  ? 

The  solution  of  this  problem  summarizes  the  solution  of  all  prob- 
lems like  those  from  5  to  10. 

12.  At  what  times  between  12  o'clock  and  6  o'clock  are  the 
hands  of  a  watch  together  ?  (Find  the  time  required  to  gain 
one  circuit,  two  circuits,  etc.) 

PROBLEMS  INVOLVING  THE  LEVER 

1.  A  teeter  board  is  in  balance  when  two  boys,  A  and  B, 
weighing  105  and  75  pounds,  respectively,  are  seated  at  dis- 
tances 5  and  7  feet  from  the  fulcrum  because  7'75  =  5-105. 
If  now  two  boys  weighing  48  and  64  pounds  are  seated  on  the 
same  board  with  the  other  boys,  the  teeter  will  again  be  in 
balance  if  their  distances  are  4  and  3  feet,  because 

7  •  75  +  4  .  48  =  5  •  105  +  3  .  64 

jB  (105  lbs.)       D (64  lbs.) (7(481bs.^  (75  lbs.)  X 

L -zrs-iz  i3-_re:erz  "_-^-_- --_-_-  rseKL" -_"_-  z.--.i J 

6  feet  ^  7  feet 

The  weight  of  the  boy  multiplied  by  his  distance  from  the 
fulcrum  is  called  his  leverage.  The  sum  of  the  leverages  on 
the  two  sides  must  be  the  same.  Hence,  if  two  boys,  weighing 
respectively,  w^  and  w^  pounds,  are  sitting  at  distances  di  and 
^2  on  one  side,  and  two  boys,  weighing  w^  and  w^  pounds,  sitting 
.    at  distances  c^g,  d^  on  the  other  side,  then 

wA  -h  w^d^  =  w^ds  +  w^.  (2) 

2.  If  two  boys  weighing  75  and  90  pounds  sit  at  distances 
of  3  and  5  feet  respectively  on  one  side  and  one  weighing 


PROBLEMS  INVOLVING  THE  LEVER        271 

82   pounds  sits   at  3   feet   on  the  other  side,  where  should 
a  boy  weighing  100  pounds  3it  to  make  the  board  balance  ? 

3.  A  beam  carries  a  weight  of  240  pounds  7^  feet  from  the 
fulcrum  and  a  weight  of  265  pounds  at  the  opposite  end  which 
is  10  feet  from  the  fulcrum.  On  which  side  and  how  far  from 
the  fulcrum  should  a.  weight  of  170  pounds  be  placed  so  as  to 
make  the  beam  balance  ? 

4.  Two  boys,  A  and  B,  having  a  50-lb.  weight  and  a  teeter 
board,  proceed  to  determine  their  respective  weights  as  fol- 
lows :  They  find  that  they  balance  when  jB  is  9  feet  and  A  7 
feet  from  the  fulcrum.  If  B  places  the  50-lb.  weight  on  the 
board  beside  him,  they  balance  when  5  is  3  and  A  4  feet  from 
the  fulcrum.     How  heavy  is  each  boy  ? 

5.  C  is  6^  feet  from  the  point  of  support  and  balances  D, 
who  is  at  an  unknown  distance  from  this  point.  C  places  a 
33-lb.  weight  beside  himself  on  the  board  and,  when  4|  feet  from 
the  fulcrum,  balances  D  who  remains  at  the  same  point  as  be- 
fore. D's  weight  is  84  pounds.  What  is  O's  weight,  and  how 
far  is  D  from  the  fulcrum  ? 

6.  E  weighs  95  pounds  and  F  110  pounds.  They  balance 
at  certain  unknown  distances  from  the  fulcrum.  E  then  takes 
a  30-pound  weight  on  the  board,  which  compels  F  to  move  3  feet 
farther  from  the  fulcrum.  How  far  from  the  fulcrum  was  each 
of  the  boys  at  first  ? 

PROBLEMS  INVOLVING  GEOMETRY 

1.  A  picture  is  4  inches  longer  than  it  is  wide.  Another 
picture,  which  is  12  inches  longer  and  6  inches  narrower,  con- 
tains the  same  number  of  square  inches.  Find  the  dimensions 
of  the  pictures. 

2.  A  picture,  not  including  the  frame, 
is  8  inches  longer  than  it  is  wide.  The 
area  of  the  frame,  which  is  2  inches  wide, 
is  176  square  inches.  Find  the  dimensions 
of  the  picture. 


a;  +  8 


272 


GENERAL  REVIEW 


3.  A  picture,  including  the  frame,  is  10  inches  longer  than 
it  is  wide.  The  area  of  the  frame,  which  is  3  inches  wide,  is 
192  square  inches.     What  are  the  dimensions  of  the  picture  ? 

'  4.  The  base  of  a  triangle  is  11  inches  greater  than  its  alti- 
tude. If  the  altitude  and  base  are  both  decreased  7  inches,  the 
area  is  decreased  119  square  inches.  Find  the  base  and  alti- 
tude of  the  triangle. 

5.  The  base  of  a  triangle  is  3  inches  less  than  its  altitude. 
If  the  altitude  and  base  are  both  increased  by  5  inches,  the 
area  is  increased  by  155  square  inches.  Find  the  base  and 
altitude  of  the  triangle. 

6.  A  square  is  inscribed  in  a  circle  and  another  circum- 
scribed about  it.  The  area  of  the  strip  inclosed  by  the  two 
squares  is  25  square  inches.     Find  the  radius  of  the  circle. 

7.  Find  the  sum  of  the  areas  of  a  circle 
of  radius  6  and  the  square  circumscribed 
about  the  circle. 

The  area  of  the  circle  is  6*  tt  =  36  tt,  and  the 
area  of  the  square  is  4  •  6^  =  4  •  36  ;  i.e.  the 
square  contains  4  squares  whose  sides  are  6. 
The  sum  of  the  areas  is 

4  .  36  +  36  TT  =  (4  -f  tt)  36  =  (4  +  3f)  36. 

8.  Find  an  expression  for  the  sum  of  the  areas  of  a  circle 
of  radius  r  and  the  circumscribed  square.  (Solve  7  by  sub- 
stituting in  the  formula  here  obtained.) 

9.  If  the  sum  of  the  areas  of  a  circle  and  the  circumscribed 
square  is  64,  find  the  radius  of  the  circle. 

By  the  formula  obtained  under  Ex.  8, 

64  =  (4  +  7r)r2  =  Y-r2. 
Hence,  r  =  Va96  =  2.99. 

10.   If  the  sum  of  the  areas  of  a  circle  and  the  circumscribed 
square  is  640.  square  feet,  find  the  radius  of  the  circle. 


PROBLEMS  INVOLVING  GEOMETRY 


273 


11.  The  sum  of  the  areas  of  a  circle  and  the  circumscribed 
square  is  a.  Find  an  expression  representing  the  radius  of 
the  circle.     (Replace  tt  by  3^  before  simplifying.) 


the   difference    between  the   areas   of  the 
circle  and  the  circumscribed  square. 

13.  If  the  radius  of  a  circle  is  r,  find  the 
difference  between  the  areas  of  the  circle 
and  the  circumscribed  square.  (Solve  Ex. 
12  by  the  use  of  the  formula  obtained  here.) 

14.  If  the  radius  of  a  circle  is  16,  find  the  area  of  the 
inscribed  square.  (This  is  the  same  problem  as  finding 
the  area  of  a  square  whose  diagonal  is  32.  See  problems 
8  and  9,  page  190.) 

15.  If  the  radius  of  a  circle  is  r,  find  an  expression  represent- 
ing the  area  of  the  inscribed  square.  (This 
is  problem  9,  page  190.) 

16.  If  the  radius  of  a  circle  is  12,  find 
the  difference  between  the  area  of  the  circle 
and  the  area  of  the  inscribed  square. 

17.  If  the  radius  of  a  circle  is  r,  find 
an  expression  representing  the  difference 

between  the  areas  of  the  circle  and  the  inscribed  square. 

18.  The  radius  of  a  circle  is  10.  Find  the  area  of  an  in- 
scribed hexagon.     (See  note,  page  192.) 

19.  The  radius  of  a  circle  is  6.  Find  the 
difference  between  the  areas  of  the  circle 
and  the  inscribed  hexagon. 

20.  Find  an  expression  representing  the 
difference  between  the  areas  of  a  circle  with 
radius  r  and  the  inscribed  regular  hexagon. 


274  GENERAL   REVIEW 

MISCELLANEOUS  PROBLEMS 

1.  Divide  the  number  645  into  two  parts,  such  that  13  times 
the  first  part  is  20  more  than  6  times  the  other. 

2.  Divide  the  number  a  into  two  parts,  such  that  h  times 
the  first  part  is  c  more  than  d  times  the  second  part. 

3.  The  sum  of  three  numbers  is  98.  The  second  is  7 
greater  than  the  first,  and  the  third  is  9  greater  than  the  sec- 
ond.    What  are  the  numbers  ? 

4.  The  sum  of  three  numbers  is  s.  The  second  is  a  greater 
than  the  first,  and  the  third  is  h  greater  than  the  second. 
What  are  the  numbers  ? 

5.  One  boy  runs  around  a  circular  track  in  26  seconds,  and 
another  in  30  seconds.  In  how  many  seconds  will  they  again 
be  together,  if  they  start  at  the  same  time  and  place  and  run 
in  the  same  direction  ? 

6.  A  bird  flying  with  the  wind  goes  &^  miles  per  hour,  and 
flying  against  a  wind  twice  as  strong  it  goes  20  mijes  per  hour. 
What  is  the  rate  of  the  wind  in  each  case  ? 

7.  A  steamer  going  with  the  tide  makes  19  miles  per  hour, 
and  going  against  a  current  \  as  strong  it  makes  13  miles  per 
hour.     What  is  the  speed  of  the  steamer  in  still  water  ? 

8.  Find  the  time  between  4  and  5  o'clock  when  the  hands 
of  the  clock  are  30  minute  spaces  apart. 

9.  A  man  takes '  out  a  life  insurance  policy  for  which  he 
pays  in  a  single  payment.  Thirteen  years  later  he  dies  and  the 
company  pays  $  12,600  to  his  estate.  It  was  found  that  his 
investment  yielded  2  %  simple  interest.  How  much  did  he 
pay  for  the  policy  ? 

10.  After  deducting  a  commission  of  3  %  for  selling  bonds, 
a  broker  forwarded  $824.50.  What  was  the  selling  price 
of    the    bonds  ? 

11.  A  broker  sold  stocks  for  $  1728  and  remitted  $  1693.44 
to  his  employer.     What  was  the  rate  of  his  commission? 


MISCELLANEOUS   PROBLEMS  275 

12.  The  difference  between  the  areas  of  a  circle  and  its  cir- 
cumscribed square  is  12  square  inches.  Find  the  radius  of 
the  circle.     (See  problem  9,  page  272.) 

13.  The  difference  between  the  areas  of  a  circle  and  its 
inscribed  square  is  12  square  inches.  Find  the  radius  of  the 
circle. 

14.  The  difference  between  the  areas  of  a  circle  and  the 
regular  inscribed  hexagon  is  12  square  inches.  Find  the  radius 
of  the  circle. 

15.  The  altitude  of  an  equilateral  triangle  is  6.  Find  its  side 
and  also  its  area.     Find  the  side  and  area,  if  the  altitude  is  h. 

16.  The  radius  of  a  circle  is  3  feet.  Find  the  area  of  the 
regular  circumscribed  hexagon.  Find  the  area  if  the  radius 
is  r  feet. 

17.  The  radius  of  a  circle  is  r.  Find  the  difference  between 
the  areas  of  the  circle  and  the  regular  circumscribed  hexagon. 

18.  The  difference  between  the  areas  of  a  circle  and  the 
regular  circumscribed  hexagon  is  9  square  inches.  Find  the 
radius  of  the  circle. 

19.  A  circle  is  inscribed  in  a  square  and  another  circum- 
scribed about  it.  The  area  of  the  ring  formed  by  the  two  cir- 
cles is  25  square  inches.     How  long  is  the  side  of  the  square  ? 

20.  In  a  building  there  are  at  work  18  carpenters,  7  plumbers, 
13  plasterers,  and  6  hod  carriers.  Each  plasterer  gets  $  1.90 
per  day  more  than  the  hod  carriers,  the  carpenters  get  35 
cents  per  day  more  than  the  plasterers,  and  the  plumbers 
50  cents  per  day  more  than  the  carpenters.  If  one  day's 
wages  of  all  the  men  amount  to  $  183.45,  how  much  does  each 
get  per  day  ? 

21.  A  train  running  46  miles  per  hour  leaves  Chicago  for 
New  York  at  7  a.m.  Another  train  running  ^6  miles  per  hour 
leaves  at  9.30  a.m.  Find  when  the  trains  will  be  15  miles 
apart.     (Two  answers.) 


276  GENERAL  REVIEW 

22.  There  is  a  number  consisting  of  three  digits,  those  in 
tens'  and  units'  places  being  the  same.  The  digit  in  hundreds' 
place  is  4  times  that  in  units'  place.  If  the  order  of  the  digits 
is  reversed,  the  number  is  decreased  by  594.  What  is  the 
number  ? 

23.  A  hound  pursuing  a  deer  gains  400  yards  in  25  minutes, 
if  the  deer  runs  1300  yards  a  minute,  how  fast  does  the  hound 
run  ?  If  the  hound  gains  Vi  yards  in  t  minutes  and  the  deer 
runs  V2  yards  per  minute,  find  the  speed  of  the  hound. 

24.  A  disabled  steamer  240  knots  from  port  is  making  only 
4  knots  an  hour.  By  wireless  telegraphy  she  signals  a  tug, 
which  comes  out  to  meet  her  at  17  knots  an  hour.  In  how 
long  a  time  will  they  meet?  If  the  steamer  is  s  knots  from 
port  and  making  Vi  knots  per  hour,  and  if  the  tug  makes  V2 
knots  per  hour,  find  how  long  before  they  will  meet. 

25.  A  motor  boat  starts  7|-  miles  behind  a  sailboat  and  runs 
11  miles  per  hour  while  the  sailboat  makes  61  miles  per  hour. 
How  far  apart  will  they  be  after  sailing  IJ  hours?  If  the 
motor  boat  starts  s  miles  behind  the  sailboat  and  runs  Vi  miles 
per  hour,  while  the  sailboat  runs  V2  miles  per  hour,  how  far 
apart  will  they  be  in  t  hours? 

26.  An  ocean  liner  making  21  knots  an  hour  leaves  port 
when  a  freight  boat  making  8  knots  an  hour  is  already  1240 
knots  out.  In  how  long  a  time  will  the  two  boats  be  280  knots 
apart?  Is  there  more  than  one  such  position?  If  the  liner 
makes  Vi  knots  per  hour  and  the  freight  boat,  which  is  Sj  knots 
out,  makes  V2  knots  per  hour,  how  long  before  they  will  be  Sg 
knots  apart? 

27.  A  passenger  train  running  45  miles  per  hour  leaves  one 
terminal  of  a  railroad  at  the  same  time  that  a  freight  running 
18  miles  per  hour  leaves  the  other.  If  the  distance  is  500  miles, 
in  how  many  hours  will  they  meet?  If  they  meet  in  8  hours, 
how  long  is  the  road?  If  the  rates  of  the  trains  are  Vi  and  v^ 
and  the  road  is  s  miles  long,  find  the  time.  ; 


FIRST  PRINCIPLES  OF  ALGEBRA 

ADVANCED    COURSE 

PART  THREE 

CHAPTER    I 
FUNDAMENTAL  LAWS 

1.  We  have  seen  in  the  Elementary  Course  that  algebra,  like 
arithmetic,  deals  with  numbers  and  with  operations  upon  num- 
bers. We  now  proceed  to  study  in  greater  detail  the  laws  that 
underlie  these  operations. 

THE   AXIOMS   OF   ADDITION   AND   SUBTRACTION 

2.  In  performing  the  elementary  operations  of  algebra  we 
assume  at  the  outset  certain  simple  statements  called  axioms. 

Definition.  Two  number  expressions  are  said  to  be  equal  if 
they  represent  the  same  number. 

Axiom  I.  If  equal  numhers  are  added  to  equal  num- 
bers, the  sums  are  equal  nuiribers. 

That  is,  if  a  =  6  and  c  —  d,  then  a  +  c  =  2>  +  rf. 

Axiom  I  implies  that  two  numbers  have  one  and  only  one  sum. 
This  fact  is  often  referred  to  as  the  uniqueness  of  addition. 

3.  If  a  =  c  and  b  =  c  then  a  =  b,  since  the  given  equations 
assert  that  a  is  the  same  number  as  b.  Hence  the  usual  state- 
ment :  If  each  of  two  numbers  is  equal  to  the  same  number,  they 
are  equal  to  each  other. 

4.  The  sum  of  two  numbers,  as  6  and  8,  may  be  found  by 
adding  6  to  8  or  8  to  6,  in  either  case  obtaining  14  as  the  result. 

277 


278  FUNDAMENTAL   LAWS 

This  is  a  particular  case  of  a  general  law  for  all  numbers  of 
algebra,  which  we  state  as 

Axiom  II.  The  sum  of  two  nuinbers  is  the  same  in  what- 
ever order  they  are  added. 

This  is  expressed"  in  symbols  by  the  identity  : 

a4-6  =  6  +  a.  [See  §  75,  E.  C.*] 

Axiom  II  states  what  is  called  the  commutative  law  of  addition, 
since  it  asserts  that  numbers  to  be  added  may  be  commuted  or 
interchanged  in  order. 

Definition.    Numbers  which  are  to  be  added  are  called  addends. 

5.  In  adding  three  numbers  such  as  5,  6,  and  7  we  first  add 
two  of  them  and  then  add  the  third  to  this  sum.  It  is  imma- 
terial whether  we  first  add  5  and  6  and  then  add  7  to  the  sum, 
or  first  add  6  and  7  and  then  add  5  to  the  sum.  This  is  a  par- 
ticular case  of  a  general  law  for  all  numbers  of  algebra,  which 
we  state  as 

Axiom  III.  The  sum  of  three  numbers  is  the  same  in 
whatever  manner  they  are  grouped. 

In  symbols  we  have        a  -\-  b  -{-  c  =  a  -{-{b  -\-  c). 

When  no  symbols  of  grouping  are  used,  we  understand  a  -\-h  -\-  c 
to  mean  that  a  and  h  are  to  be  added  first  and  then  c  is  to  be  added  to 
the  sum. 

Axiom  III  states  what  is  called  the  associative  law  of  addition, 
since  it  asserts  that  addends  may  be  associated  or  grouped  in 
any  desired  manner. 

It  is  to  be  noted  that  an  equality  may  be  read  in  either  direction. 
Thus        a  +  h  +  c  =  a-\-(h  +  c)  and  a+(&  +  c)=a+&  +  c 
are  equivalent  statements. 

6.  If  any  two  numbers,  such  as  19  and  25,  are  given,  then 
in  arithmetic  we  can  always  find  a  number  which  added  to 
the  smaller  gives  the  larger  as  a  sum.  That  is,  we  can  sub- 
tract the  smaller  number  from  the  larger. 

*  E.  C.  means  the  Elementary  Course. 


AXIOMS  279 

In  algebra,*  where  negative  numbers  are  used,  any  number 
may  be  subtracted  from  any  other  number. 

That  is :  For  any  pair  of  numbers  a  and  b  there  is  one  and 
only  one  number  c  such  that  0  +  0  =  b. 

The  process  of  finding  the  number  c  when  a  and  b  are  given 
is  called  subtraction.  This  operation  is  indicated  thus,  b—a=c, 
where  b  is  the  minuend,  a  the  subtrahend,  and  c  the  remainder. 

Since  for  a  given  minuend  and  a  given  subtrahend  there  is 
one  and  only  one  remainder,  we  have  for  all  numbers  of 
Algebra, 

Axiom  IV.  If  equal  numbers  are  subtracted  from  equal 
numbers,  the  remainders  are  equal  numbers. 

Definitions.  If  a  -f-  c  =  a,  then  the  number  c  is  called  zero, 
and  is  written  0.^  That  is,  a  +  0  =  a,  or  a  — a  =  0.  Hence 
zero  is  the  remainder  when  minuend  and  subtrahend  are  equal. 

By  definition  of  subtraction,  the  equality  b  —  a  =  c  implies 
that  c  is  a  number  such  that  c-{-a  =  b. 

A  dding  a  to  each  member  of  the  equality  b  —  a  =  c,  we  have 
6  —  a  +  a  =  c4-a,  which  by  hypothesis  is  equal  to  b.  Hence 
subtracting  a  number  and  then  adding  the  same  number  gives  as 
a  result  the  original  number  operated  upon. 

Axiom  IV  implies  the  uniqueness  of  subtraction. 

THE  AXIOMS  OF  MULTIPLICATION  AND  DIVISION 

7.  Axioms  similar  to  those  just  given  for  addition  and  sub- 
traction hold  for  multiplication  and  division. 

Axiom  V.  If  equal  nurribers  are  multiplied  by  equal 
numbers,  the  products  are  equal  numbers. 

This  axiom  implies  the  uniqueness  of  multiplication.  That  is, 
two  numbers  have  one  and  only  one  product. 

8.  The  product  of  5  and  6  may  be  obtained  by  taking  5  six 
times,  or  by  taking  6  five  times.-  That  is,  b  'Q  =  &  -5.  This 
is  a  special  case  of  a  general  law  for  all  numbers  of  algebra, 
which  we  state  as 


^80  FUNDAMENTAL  LAWS 

Axiom  VI.    The  product  of  two  numbers  is  the  same  in 
whatever  order  they  are  multiplied. 
In  sjanbols  we  have        a  >  b  =  b  •  a. 

This  axiom  states  what  is  called  the  commutative  law  of 
factors  in  multiplication. 

9.  The  product  of  three  numbers,  such  as  5,  6,  and  7,  may 
be  obtained  by  multiplying  5  and  6,  and  this  product  by  7,  or  6 
and  7,  and  this  product  by  5.  This  is  a  special  case  of  a  gen- 
eral law  for  all  numbers  of  algebra,  which  we  state  as 

Axiom  VII.  The  product  of  three  numhers  is  the  same 
in  whatever  ^manner  they  are  grouped. 

In  symbols  \ye  have      abc  —  a  (be). 

The  expression  abc  without  symbols  of  grouping  is  understood  to 
mean  that  the  product  of  a  and  b  is  to  be  multiplied  by  c. 

This  axiom  states  what  is  called  the  associative  law  of  factors 
in  multiplication. 

Principles  IV  and  XIII  of  E.  C.  follow  from  Axioms  VT  and  VII. 

10.  Another  law  for  all  numbers  of  algebra  is  stated  as 

Axiom  VIII.  The  product  of  the  sum  or  difference  of  two 
numbers  and  a  given  number  is  equal  to  the  result  obtained 
by  multiplying  each  number  separately  by  the  given  num- 
ber and  then  adding  or  subtracting  the  products. 

In  symbols  we  have 

a{b  +  c)  =  ab -\- ac  and  a(b  —  c)=ab  —  ac. 

Axiom  VIII  states  what  is  called  the  distributive  law  of 
multiplication. 

When  these  identities  awe  read  from  left  to  right,  they  are  equivar 
lent  to  Principle  II,  E.  C,  and  when  read  from  right  to  left  (see 
§  5)  they  are  equivalent  to  Principle  I,  E.  C.  The  form  a  (6  ±  c)  = 
ab  ±  ac  is  directly  applicable  to  th'e  multiplication  of  a  polynomial  by 
a  monomial,  and  the  form  ab  ±ac  =  a(b  ±  c),  to  the  addition  and 
subtraction  of  monomials  having  a  common  factor. 


AXIOMS  281 

11.  Definitions.  If  ac  —-  b,  the  process  of  finding  c  when  a 
and  b  are  given  is  called  division.     This  operation  is  indicated 

thus :  b-i-a  =  c,  01-  -  =  c,  where  b  is  the  dividend,  a  thfi  divisor, 

and  c  the  quotient.     For  the  case  a  =  0  see  §§  24,  25. 

Axiom  IX.  If  equal  numbers  are  divided  hy  equal  nuin- 
bers  {the  divisors  being  different  from  zero)  the  quotients 
are  equal  numbers. 

Definition.   If  a-  c  =  a,  a^ 0,*  then  the  number  c  is  called 

unity,  and  is  written  1.     That  is,  -  =  1.     Hence  unity  is  the 

a 

quotient  when  dividend  and  divisor  .are  equal. 

By  definition  of  division,  the  equality  -  =  c  implies  that  c  is 
a  number  such  that  ac^  b. 

Multiplying  both  sides  of  the  equality  -  =  c  by  a,  we  have 

a-  -  =  ac,  which  by  hypothesis  equals  b.     Hence  dividing  by  a 

number  and  then  multiplying  by  the  same  number  gives  as  a  result 
the  original  number  operated  upon. 

Axiom  IX  implies  the  uniqueness  of  division.  That  is :  For 
any  two  numbersj  a  and  b,  a  ^  0,  there  is  one  and  only  one  num- 
ber c  such  that  ac  =  bj  or  -  =  c. 

a  . 

12.  Axioms  I,  IV  (in  case  the  subtrahend  is  not  greater  than 
the  minuend),  V,  and  IX  underlie  re'spectively  the  processes 
of  addition,  subtraction,  multiplication,  and  division,  from  the 
very  beginning  in  elementary  arithmetic.  Axioms  II,  III,  VI, 
VII,  and  VIII  are  also  fundamental  in  arithmetic,  where  they 
are  usually  assumed  without  formal  statement. 

E.g.  Axiom  VIII  is  used  in  long  multiplication,  such  as  125  x  235, 
where  we  multiply  125  by  5,  by  30,  and  by  200,  and  then  add  the 
products. 

*  The  symbol  a  =5^  0  stands  for  the  expression  a  is  not  equal  to  zero. 


282  FUNDAMENTAL   LAWS 

13.  Negative  Numbers.  Axiom  IV,  in  case  the  subtrahend  is 
greater  than  the  minuend,  does  not  hold  in  arithmetic  because 
of  the  absence  of  the  negative  number.  This  axiom  therefore 
brings  the  negative  number  into  algebra. 

We  now  proceed  to  study  the  laws  of  operation  upon  this 
enlarged  number  system.  In  the  Elementary  Course  concrete 
applications  were  used  to  show  that  certain  rules  of  signs  hold 
in  operations  upon  positive  and  negative  numbers.  The  same 
rules  follow  from  the  axioms  just  stated. 

14.  Definitions.  If  a  +  6  =  0,  then  b  is  said  to  be  the  negative 
of  a  and  a  the  negative  of  b.  If  a  is  a  positive  number,  that 
is,  an  ordinary  number  of  arithmetic,  then  b  is  called  a  negative 
number.  We  denote  the'  negative  of  a  by  —  a.  Hence, 
a  +  (—  a)  =  0.     a  and  —  a  have  the  same  absolute  value. 

If  a  —  b  is  positive,  then  a  is  said  to  be  greater  than  b. 
This  is  written  a>b.  li  a  —  b  is  negative,  then  a  is  said 
to  be  less  than  b.     This  is  written  a<b.     If  a  —  6  =  0,  then 

a  =  b.     See  §  6. 

« 

PRINCIPLES   OF   ADDITION   AND   SUBTRACTION 

15.  We  now  show  that 

a4.(-A)=a-6.  See  §  51,  E.  C. 

Let  a +  (-&)  =  a,-.  (1) 

Adding  b  to  ea,(^  member  of  this  equation  we  have,  by  Axiom  I, 

a  +  (-b)-\-b  =  x-i-h.  (2) 

But  by  the  associative  law, 

a  +  (-  &)  +  ft  =  a  +  [(-  6)  4-  ft]  =  a.  (3) 

Hence,  a  =  a:  +  ft,  or  (§  6),  a  -  ft  =  x.  (4) 

From  (l)'and  (4),        a  +  (-  h)  =  a  -  b. 

That  is :  Adding  a  negative  number  is  equivalent  to  subtracting 
this  number  with  its  sign  changed. 

It  follows  that  either  of  the  symbols,  +(—b)  and  —  b,  may 
replace  the  other  in  any  algebraic  expression. 


PRINCIPLES  288 

16.  It  is  an  immediate  consequence  of  §  15  that  a  parenthesis 
preceded  by  the  plus  sign  may  he  removed  without  changing  the 
sign  of  any  term  within  it.     See  §  79,  E.  C. 

It  also  follows  that  an  expression  may  he  inclosed  in  a  paren- 
thesis preceded  hy  the  plus  sign  without  changing  the  sign  of  any 
of  its  terms. 

17.  To  show  that  a-{-b)  =  a  +  b.  See  §  57,  E. C. 
Let  a-{-h)=x.  (1) 
Adding  (  — Z>)  to  both  members  (Ax.  I), 

a-{-h)^-{-h)  =  x-\-{-h)=x-b.  (2) 

But  a-{-b)  +  {-h)=a.  (3) 

Hence,         a  =  x  —  hora-\-i  =  x.  (4) 

From  (1)  and  (4)  we  have  a  —  (—  6)  —  a  +  6. 
That  is :  Subtracting  a  negative  number  is  equivalent  to  adding 

this  number  with  its  sign  changed. 

It  follows  that  either  of  the  symbols  —(—b)  and  +6  may 

replace  the  other  in  any  algebraic  expression. 

18.  To  show  that 

a-(b--c-^d)=a—  b+c-d.     See  §  79,  E.  C. 
Let  a-{h  -cJcd)=x.  (1) 

Then  a  =  x  -\-{b  -  c  -\-  d)=  x  ^- h  -  c  +  d.  (2) 

Adding  c  and  subtracting  b  and  d  from  each  member, 
we  have  a  —  b-\-c  —  d  =  x.  (3) 

From  (1)  and  (3),  a- {b  -c  +  d)=  a -h  -{■  c  -  d.  (4) 

That  is :  A  parenthesis  preceded  by  the  minus  sign  may  be  re- 
moved by  changing  the  sign  of  each  term  within  it. 

It  also  follows  from  equation  (4),  read  from  right  to  left, 
that  an  expression  may  be  inclosed  in  a  parenthesis  preceded  by  a 
minus  sign,  if  the  sign  of  each  term  within  is  changed, 

19.  It  follows  by  use  of  §  18  that 

a-b  =  -(b-a). 
For  a  —  b  =  —  b-\-a  =  ~(b  —  a). 


284  FUNDAMENTAL  LAWS 

20.  It  follows  further  by  use  of  §  18  that 

-a+(-A)  =  -(a  +  A). 
For  ^a-{-(—b)  =  —  a—b  =  —  (a  +  b). 

21.  From  the  identities 

a-\-(-b)  =  a-b,  §15, 

a-{-b)=a-\-b,^  §17, 

a-b=-(b-a),  §19, 

^a4-(-a)  =  -(a  +  &),  §20, 

it  follows  that  addition  and  subtraction  of  positive  and  nega- 
tive numbers  are  reducible  to  these  operations  as  found  in  arith- 
metic, where  all  numbers  added  and  subtracted  are  positive, 
and  where  the  subtrahend  is  never  greater  than  the  minuend. 

E.g.  5+(-8)=5-8  =  -(8-5)  =  -3. 

5-(-8)z.5  +  8  =  13. 
-5-8  =  -(5  +  8)zz:_13. 

PRINCIPLES    OF   MULTIPLICATION   AND   DIVISION 

22.  To  show  that  a  -0  orO  -  a  equals  0  for  all  values  of  a. 
By  definition  of  zero,  a  •  0  =  a(6  —  &). 

By  Axiom  VIII,  a{h-h)=ah-  ah. 

By  definition  of  zero,      ah  —  ah  =  0. 

Hence,  a  •  0  ==  0. 

By  Axiom  VII,  a  •  0  =  0  •  a  =  0. 

It  follows  that  a  product  is  zero  if  any  one  of  its  factors  is 
zero ;  and  if  a  product  is  zero,  then  at  least  one  of  its  factors 
must  be  zero. 

23.  To  show  that  -  =  0,  provided  a  is  not  zero. 

Since  by  §  22,  0  =  a  •  0,  we  have  by  the  definition  of  division  -  =  0. 

a 

24.  To  show  that  -  represents  any  number  whatever. 

0 

That  is,  ji  —  l^i  for  all  values  ofk. 


PRINCIPLES  285 

Since  by  §  22,  0  =  0  •  A:,  we  have  by  the  definition  of  division  -  =  ife 

0  ^ 

for  all  values  of  k.     Hence,  -  does  not  represent  any  definite  number. 

25.  To  show  that  there  is  no  numher  k  such  that  ~  =  k, 
provided  a  is  not  zero. 

If  -  =  ^,  then  by  definition  of  division,  /: .  0  =  a.      But  by  §  22, 

^  •  0  =  0  f or  all  values  of  k.     Hence,  if  a  is  not  zero,  k  is  impossible. 

From  §§  24,  25,  it  follows  that  division  by  zero  is  to  be  ruled 
out  in  all  cases  unless  special  interpretation  is  given  to  the 
results  thus  obtained. 

26.  To  show  that  a  (-/&)  =  -  a6.     See  §  63,  E.  C. 

Let  a(-b)  =  x.  (1) 

Adding  ab  to  both  members, 

a(-b)-{-ab  =  x  +  ab.  (2) 

By  Axiom  VIII,      al(- b)-\-b']  =x +  ab,  '  (3) 

or,  a  .  0  =  0  =  z  +  a&.  (4) 

Hence  (§14),  x  =  -ab.  (5) 

From  (1)  and  (5),  a(-b)=-  ab. 

That  is,  the  product  of  a  positive  and  a  negative  number  i^ 
negative. 

27.  To  show  that  {-a){-b)=ab.     See  §  63,  E,  C. 

Let                                  (_a)(_6)  =  x.  (1) 
Adding  (—  a)b  to  each  member, 

i-a){-b)  +  {-a)b  =  x  +  (^-a)b  =  x-ab.  (2) 

By  Axiom  VIII,      (  _  a)  [(  -  ft)  +  6]  =  a:  -  a&,  (3) 

or,                                                              Q  =  x-ab.  (4) 

Hence  (§14),                                    ab  =  x.  .                (5) 

From  (1)  and  (5),  {~-a){-b)  =  ab. 

That  is,  the  product  of  two  negative  numbers  is  positive. 


286  FUNDAMENTAL  LAWS 

28.  To  show  that  if  the  signs  of  the  dividend  and  divisor 
are  alike,  the  quotient  is  positive;  and  if  unlike,  the  quo- 
tient is  negative.    See  §  ijQ,  E.  C. 

*  If  a  =  be,  then  by  the  law  pf  signs  in  multiplication,  —  a  =(—  6)c, 
—  a  =  6(—  c),  and  a  =  (—  Z>)(—  c).  Hence,  by  definition  of  division, 
we  have  respectively : 

a  —a  —a  ,    a 

-  =  c, =  c,    -—  =  -c,    and  — r  =  -c. 

b  —b  b  —b 


29.   To  show  that 

G                 C 

Let 

x  =  ^-b. 
c 

(1) 

Then 

cx  =  c  '-•b=ab. 
c 

(2) 

Dividing  by  c, 

ab 
x  =  — 
c 

(3) 

From  (1)  and  (3), 

a    y  _ab 
c            c 

(4) 

30.   To  show  that 

a-\-ba  .  b 
c         c      c 

By  §  29, 

a  +  b^l.{a-^b)^l 
c                 c             c 

(1) 

By  Axiom  VIII, 

c                  c            c           c      c 

(2) 

Hence, 

a  +  b  _a      h 
c         c      c 

(3) 

31.   To  show  that 

a  —  b     a      b 
c         c     c 

By  §  29 

c                  c             c^ 

(1) 

By  Axiom  VIII,      i( 

7N      -1            1    y       a      b 

a  —  b)  =-  -a  —  'b  = 

c            e            c     c 

(2) 

Hence, 

a—b      a      b 
c          c      c 

(3) 

CHAPTER   II 
FUNDAMENTAL    OPERATIONS 

32.  The  operations  of  addition,  subtraction,  multiplication, 
division,  and  finding  powers  and  roots  are  called  algebraic 
operations. 

33.  An  algebraic  expression  is  any  combination  of  number 
symbols  (Arabic  figures  or  letters  or  both)  by  means  of  indi- 
cated algebraic  operations. 

E.g.  24,  3  +  7,  9(&  4-  c),    ^LilJ^,  x"^  +  y/y,  are  algebraic  expressions. 

34.  Any  number  symbol  upon  which  an  algebraic  operation 
is  to  be  performed  is  called  an  operand. 

All  the  algebraic  operations  have  been  used  in  the  Elementary 
Course.  They  are  now  to  be  considered  in  connection  with  the  fun- 
damental laws  developed  in  the  preceding  chapter,  and  then  applied 
to  more  complicated  expressions.  The  finding  of  powers  and  roots 
will  be  extended  to  higher  cases. 

35.  One  of  the  two  equal  factors  of  an  expression  is  called 
the  square  root  of  the  expression ;  one  of  the  three  equal  fac- 
tors is  called  its  cube  root;  one  of  the  four  equal  factors,  its 
fourth  root,  etc.  A  root  is  indicated  by  the  radical  sign  and  a 
number,  called  the  index  of  the  root,  which  is  written  within 
the  sign.     In  the  case  of  the  square  root,  the  index  is  omitted. 

E.g.  Vi  is  read  the  square  root  ofi  ;  y/8  is  read  the  cube  root  ofS; 
y/M  is  read  the  fourth  root  of  64,  etc. 

36.  A  root  which  can  be  expressed  in  the  form  of  an  integer, 
or  as  the  quotient  of  two  integers,  is  said  to  be  rational,  while 
one  which  cannot  be  so  expressed  is  irrational. 

E.g.  \^  =  2,  Va2  +  2ah  +  b^  =  a  +  b,  and  \/|  =  f  are  rational  roots, 
while  y/i  and  Va^  +  a6  +  &^  are  irrational  roots. 

287 


288  FUNDAMENTAL   OPERATIONS 

An  algebraic  expression  which  involves  a  letter  in  an  ir- 
rational root  is  said  to  be  irrational  with  respect  to  that  letter; 
otherwise  the  expression  is  rational  with  respect  to  the  letter. 

E.g.  a  +  bVc  is  rational  with  respect  to  a  and  h,  and  irrational 
with  respect  to  c. 

37.  An  expression  is  fractional  with  respect  to  a  given  letter 
if  after  reducing  its  fractions  to  their  lowest  terms  the  letter 
is  still  contained  in  a  denominator. 

E.g.    — ^ 1-  6  is  fractional  with  respect  to  c  and  d,  but  not  with 

c  -{•  d 
respect  to  a  and  b. 

38.  Order  of  Algebraic  Operations.  In  a  series  of  indicated 
operations  where  no  parentheses  or  other  symbols  of  aggrega- 
tion occur,  it  is  an  established  usage  that  the  operations  of 
finding  powers  and  roots  are  to  be  performed  first,  then  the 
operations  of  multiplication  and  division,  and  finally  the  opera- 
tions of  addition  and  subtraction. 

^.^.2  +  3.4  +  5.  ^8-42-8  =  2  +  3.4  +  5.2-16-4-8 

=  2  +  12  +  10-2  =  22. 

In  cases  where  it  is  necessary  to  distinguish  whether  multipli- 
cation or  division  is  to  be  performed  first,  parentheses  are  used. 

E.g.  In  6  ^  3  X  2,  if  the  division  comes  first,  it  is  written  (6  ^3)  x 
2  =  4,  and  if  the  multiplication  come  first,  it  is  written  6  -4-  (3  x  2)  =  1. 

ADDITION  AND  SUBTRACTION  OP  MONOMIALS 

39.  In  accordance  with  §  10,  the  sum  (or  difference)  of  terms 
which  are  similar*  with  respect  to  a  common  factor  (§  74,  E.  C.) 
is  equal  to  the  product  of  this  common  factor  and  the  sum  (or 
difference)  of  its  coefficients. 

Ex.  1.   Sax^  +  9  ax^  _  3  aa;2  ^  (8  +  9  -  3)  ax^  =  14  ax^. 
Ex.  2.  aVx^  +  y2  +  hVx^  +  y^  =  (a  +  b)  Vx^  +  y^. 


Ex. 


3    x(x  -  l)(x  -2)      x(x-l)^(x-0      .\x(x  -  1) 
1-2.3  1-2  \    3  I      1-2 


_x+l     x(x  -  1)  _  (re  +  l)x(x  -  1) 
3     *      1-2  1.2-3 


ADDITION  AND  SUBTRACTION  289 

EXERCISES 

Perform  the  following  indicated  operations : 


2.  3V«'-4~2VV-4  +  2V^2-r4-4\/^^'=T. 

3.  ab^c*  -  db'^c* -\- eb^c*  +  fb^c\ 

4.  aV  +  5aV  —  5aV  —  3aV. 

5.  7a^f-^5a^y*  —  9xy+5a^y*. 

6.  2a'*  +  a''-^  +  a"+^  =  a''-X2a  +  1  +  a?)  =  a^-^l  +  a)^ 

7.  7i(n  -l)(n-  2)(n  _  3)(n  -  4)  +  n(7i  -  l)(w  -  2)(n  -  3). 

n(n  -  l)(n  —  2)  (n  -  3)  is  the  common  factor  and  n  -  4  and  1  are 
the  coefficients  to  be  added. 

8.  n(n  -l)(n-  2)(?i  -  3)  (n  -  4)(n  -  5)(n  -  6) 

+  n{n  -  l)(n  -  2)(n  -  3)(w  -  4Xn  ~  5). 

9.  n{n  -  l)(n  -  2)(n  -  3)  +  (n  -  l)(?i  -  2)(w  -  3). 

10.  n{n  -  l)(n  -  2)(n  -  3)(n  -  4)  +  (n  -  l)(7i  -  2)(n  -  3). 

11.  (a  -  4)(6  +  3)  +  (2  a  - 1)(6  _  2)  +  (a  +  3)(6  +  3), 
First  add  (a  -  4)(6  +  3)  and  (a  +  3)(&  +  3). 

12.  {x  +  2y)(x  -  2y)  +  (x-  3y){x  -  2y)  +  (2x  -  2^)(a;  -  y). 

13.  (5a  -  36)(a  -  6)(a  +  &)  +  (26  -  4a)(a  -  b){a  +  b) 

+  (a-  bf(2a  -  b). 

14.  (7ar^  +  3/)(5a;  -  y){x  +  y)  +  (Ta:^  ^  g^s^^^  +  y)(2y-  Ax) 

+  (7a^-3i/^(aj  +  2/)'. 

15.  23.32.5  +  2^.3.5. 

The  common  factor  is  2^ .  3  •  5.     Hence  the  sum  is 
28.3-5(3  +2)  =  28.3.52. 

16.  2  .  3^  •  7  +  22 .  33 .  72  -  2^  •  33 . 7. 

17.  3*  .  5^ .  13  +  3^  •  5^  •  132. 


290  FUNDAMENTAL   OPERATIONS 

18.  5^  .  73  •  11  +  53  .  72  .  11  -  23  .  3  .  53  .  72  .  11. 

19.  322 .  7^»  .  131^  +  321  .  71^  .  13^^  4-  324 .  71^  .  13^^ 

20.  1.2.3...W   +1  .2.3...n(w4-l). 

The  dots  mean  that  the  factors  are  to  run  on  in  the  manner  indi- 
cated up  to  the  number  n.  The  common  factor  in  this  case  is 
1  •  2  •  3  •••  n,  and  the  coefficients  to  be  added  are  1  and  ii  -\-  1.  Hence 
the  sum  is  1  •  2  •  3  ...  n(n  +  2). 

21.  1  .  2  .  3  •••  w  -f  1  .  2  .  3  ...  ri(?i  +  1) 

+  l-2.3...w(n  +  l)(7i  +  2). 

22.  1  •  2  .  3  ...  w  +  3  .  4  .  5  ...  71  4-  5  .  6  •  7  ...  n. 

23.  n(n  -  1)  ...  (n  -  6)    +  n(n  _  1)  ...  (n  -  6)(n  -  7). 

24.  n{7i  —  1)  ...  (n  —  r)    +  n(n  —  1)  •••  (ri  —  r)(n  —  r  —  1). 

25.  wa"6  4-a."6.  26.    ^^^  ~  ^)  a»-^6^  +  yia"-^6l 

7i(yL  -  1)(^  -  2)  ^n-2^,,n(n  -  1)        2^,3 
1.2.3  1-2 

The  common  factor  is  ^^^  ~ — ^  a'^-%^  and  the  coefficients  to  be 
added  are  — - —  and  1. 


28    ^(^  -  ^)(n  -  2)(n  -  3)       3.4      n(n  -  l)(n  -  2)  ^^.^^^ 
2.3.4  2.3 


29    n(7i  ~  l)(n  -  2)(n  -  3)(yi  -  4)       ,,, 
2.3.4.5 


n(n-l)(r.-2Xn-3)       455 
^  2.3.4 


30    n(n  - 1)  ...  (n  -  r  +  l)(yi  -  r)    n-r^r+i 
2.3...r(r  +  l) 


.  n(n-l)  ...  (n-r  4-  1)  ^n-r^r+i 
2.3...r 


ADDITION   AND   SUBTRACTION  291 

ADDITION  AND  SUBTRACTION  OP  POLYNOMIALS 

40.   The  addition  of  polynomials  is  illustrated  by  the  follow- 
ing example. 
Add  2a  +  3&-4cand3a-26H-5c. 

The  sum  may  be  written  thus : 

(2a +  36 -4c)  +  (3a -26  + 5c). 

By  the  associative  law,  §  5,  and  by  §  16,  we  have, 
2a  +  36-4c  +  3a-2  6+5c. 

By  the  commutative  law,  §  4,  and  by  §  15,  this  becomes, 
2a  +  3a  +  36-2  6-4c  +  5c. 

Again  by  the  associative  law,  combining  similar  terms,  we  have, 
5  a  +  6  +  c. 

From  this  example  it  is  evident  that  several  polynomials  may 
be  added  by  combining  similar  terms  and  then  indicating  the 
sum  of  these  results. 

For  this  purpose  the  polynomials  are  conveniently  arranged  so  that 

similar  terms  shall  be  in   the   same  column.     Thus,  in  the   above 

example, 

^  2a+36-4c 

3a-26+5c 


5  a  +     6  +     c 


41.  For  subtraction  the  terms  of  the  polynomials  are  arranged 
as  for  addition.  The  subtraction  itself  is  then  performed  as  in 
the  case  of  monomials.     See  §§  17-19. 

Example.     Subtract  ^x  —  2y-\-Qz  from  3x-\-Qy  —  ^z. 

3a:  +  6y  -  3s 
4a;-2y +  62 

The  steps  are :        •         "*  V        *' 

3a;-4a;  =  -a;;   6y -(-27/)  =  8y;    -  32  -  (  + 6z)  =  -  9«. 


292  FUNDAMENTAL  OPERATIONS 

EXERCISES 

1.  Add  8a^-llx-7x%  2x-ex'+10,  -5  +  4.x^  +  9x, 
and  13  x^- 5 -12  x^. 

2.  Add  5a^-2a-12-10a%  U-7 a  +  a'-da^,  Sa' 
_13a3-f  4-lla,  aiid3-7a  +  10a2  +  4al 

3.  Erom  the  sum  of  9m^  — Sm'^  +  ^m  — 7  and  3m^  — 4  m* 

-{-2m-\-S  subtract  4m^  — 2m^  — 4  +  8m. 

4.  From  the  sum  of  x'^  —  ax^  —  aV  —  a^x  -\-2a^  and  3  a^ 
+7  a^ajs  _  5  a^a;  +  2  ci^  subtract  3  a;*  +  aa;^  -  3  aV  +  a^x  -  a\ 

5.  Add  37a-46-17c  +  15d-6/-87i  and  3c-31a 
H-96-5c?-/i-4/. 

6.  Add  llg-10_p-8w  +  3m,  24.m-17 q-{-l^p-13n, 
9n  —  6m  —  4:q  —  7p  —  5n,  and  8  g  —  4p  — 12 m  +  18 n. 

7.  From  the  sum  of  13a  — 15  6  — 7c  — lid  and  7a  — 66 
+  8 c 4- 3 d  subtract  the  sum  of  6d  —  56  —  7cH-2a  and  5 c 
-10c2-286  +  17a. 

8.  Add  23  .  3^  a^  -  2^  .  32  x2  +  22  .  33 .  7a?  +  22  .  32  .  5, 
2«  .  3-W-2*  .  32  .  7a;  +  2*  •  33ar^-22  .  3^  .  52,  and  2^  .  3^3^ 
-23.33.5  +  23.3«a;-2*.3^a;2^ 

9.  Add    (a  +  6  —  c)  m  +  (a  —  6  +  c)  n  +  (a  —  6  —  c)  /c, 

(2a-36  +  c)m  +  (6-3a  +  c)n+(4c4-26  +  a)k, 
and  (6  -  2  c)m  +(2  a  -  2  c  +  6)n  4-(2  6  -  2  a  +  c) A:. 

10.  From  the  sum  of  aa^  —  bx^  +  cx—d  and  6a^  +  aa;^  —  dx 
^c  subtract  (a  — 6)a^  H-(c  — a)aj2  — (6-|- d)a;  — d  +  c. 

11.  From  (m  —  n)(m  —  n)a^  +  (n  —  m^x^  —  (n  +  m)  a?  4-  8  sub- 
tract the  sum  of  n  (m  —  n)  a.*^  —  4  (n  —  m)  V  +  (n  +  m)  a;  —  31 
and  2(n  —  m) V  —  m(m  —  71)3?  —  2  (n  +  m)  a;  -|-  25. 

12.  Add  a**  4- 2  a"+^  +  a"+2  and  2  a"  -  4  a"+^  +  5  a«+2  and 
from  this  sum  subtract  7  a'*+^  —  80**  +  a"+2. 


ADDITION  AND   SUBTRACTION  293 

REMOVAL   OF  PARENTHESES 

42.  By  the  principles  of  §§  15-18,  a  parenthesis  inclosing  a 
polynomial  may  be  removed  with  or  without  the  change  of 
sign  of  each  terra  included,  according  as  the  sign  — .  or  +  pre- 
cedes the  parenthesis. 

In  case  an  expression  contains  signs  of  aggregation,  one 
within  another,  these  may  be  removed  one  at  a  time,  beginning 
with  the  innermost,  as  in  the  following  example : 

a-{h  +  c  -Id  -  e  +f-{g  -h)-]] 
=  a-{h-^c-ld-e  +/-  g  +  K]} 
=  a-{6  +  c-rf  +  c  -/+  g  -h} 
=  a -b  -  c  +  d  -  e -\-f- g  +  h. 

Such  involved  signs  of  aggregation  may  also  be  removed  aM 
at  once,  beginning  with  the  outermost,  by  observing  the  number 
of  minus  signs  which  affect  each  term,  and  calling  the  sign  of 
any  term  +  if  this  number  is  even,  —  if  this  number  is  odd. 

Thus,  in  the  above  example,  b  and  c  are  each  affected  by  one 
minus  sign,  namely,  the  one  preceding  the  brace.  Hence  we  write, 
a  —  b  —  c. 

d  and/ are  each  affected  by  two  minus  signs,  namely  the  one  before 
the  brace  and  the  one  before  the  bracket,  while  e  is  affected  by  these 
two,  and  also  by  the  one  preceding  it.     Hence  we  write,  d  —  e  +  f. 

g  is  affected  by  the  minus  signs  before  the  bracket,  the  brace, 
and  the  parenthesis,  an  odd  number,  while  Ti  is  affected  by  these 
and  also  by  the  one  preceding  it,  an  even  number.  Hence  we  write 
-g  +  h. 

By  counting  in  this  manner  as  we  proceed  from  left  to  right,  we 
give  the  final  form  at  once,  a  —  b  —  c  -\-  d  —  e  +/-  g  -\-h. 

EXERCISES 

In  removing  the  signs  of  aggregation  in  the  following,  either 
process  just  explained  may  be  used.  The  second  method  is 
shorter  and  should  be  easily  followed  after  a  little  practice. 

1.  7-S-4-(4-[-7])-(5-[4-5]  +  2)j. 


294  FUNDAMENTAL   OPERATIONS 

2.  _[_(7-J_4  +  9j-13)-(12-3+[-7  +  2])]. 

3.  6-(-3-[-5  +  4]+57-3-(7-19)i  +  8). 

4.  5  +  [-(-{-5-3  +  llJ-15)-3]+8. 

5.  4:X  —  [Sx  —  y  —  \Sx  —  y  —  (x  —  y  —  x)-{-x\  —  Sy^. 

The  vinculum  above  y  —  x  has  the  same  eifect  as  a  parenthesis,  i.e. 
~  y  -  x=  -  (y~  x). 

6.  Sx'-2y'-(4.x'-\Sx'-(y^-2x^-3y'l-y'  +  4:a^. 


7.  7a  — f3a  — [— 2a  — tt4-3  +  a]— 2a  — 5j. 

8.  Z-(-2m-n-5Z-mJ)-(5Z-2w- [-3m  +  n]). 

9.  2d-[3d+\2d-(e-5d)l-(d  +  Se)^. 

10.  42/-(-22/-[-32/-J-2/-^^^K22/]).       ' 

11.  3x-[Sx-(x-S)-\-2x  +  6-Sx-ll}. 


12.  aj-(a;-S-4a;-[5a;  — 2a;-5]-[-a;  — a;-3];). 

13.  3a;-5i/-[32/  +  22]-(4a;-[22/-32;]-32/-2  2)4-4a;|. 

14.  x-(-x  —  \-Sx-lx-2x-\-5^-4:}-l2x-x  —  S']). 


MULTIPLICATION  OF  MONOMIALS 

43.  In  the  elementary  course  we  saw  that  2*  •  2"  =  2*+". 
More  generally,  if  b  is  any  number  and  k  and  n  any  positive 
integers,  we  have 

bf*  '  b"  =  A*+". 

For  by  the  definition  of  a  positive  integral  exponent, 
b  =  b  -b  'b  '••  to  k  factors, 
and  b"*  =  b  'b  -b  '-'to  n  factors. 

Then,  6*  .&»»  =  (&.  6  ...  to  k  factors) (6  •  &  ...  to  n  factors) 

=  6  .  &  .  6  ...  to  ^  +  n  factors. 
Hence,  h^'b»  =  6*+~. 

That  is :  Tlie  prgduct  of  two  powers  of  the  same  base  is  a 
power  of  that  base  whose  exponent  is  the  sum  of  the  exponents 
of  the  common  ha^se. 


MULTIPLICATION   AND   DIVISION  295 

44.  In  finding  the  product  of  two  monomials,  the  factors  may 
be  arranged  and  associated  in  any  manner,  according  to  §§  8,  9. 

E.g.    (3a62)x(5a263)  =  3a&'^.5a268  §9 

=  3  .  5  .  a  .  a2  .  &2  .  ft8  §8 

=  (3.o)(a.a2)(62.68)  §9 

=  15  a%^  §  43 

The  factors  in  the  product  are  afl*ranged  so  as  to  associate  those 
consisting  of  Arabic  figures  and  also  those  which  are  powers  of  the 
«ame  base.  This  arrangement  and  association  of  the  factors  is  equiv- 
alent to  multiplying  either  monomial  by  the  factors  of  the  other  in 
succession.     See  §  134,  E.  C. 

45.  It  is  readily  seen  that  a  product  is  negative  when  it  con- 
tains an  odd  number  of  negative  factors ;  otherwise  it  is  positive. 

For  by  the  commutative  and  associative  laws  of  factors  the  negative 
factors  may  be  grouped  in  pairs,  each  pair  giving  a  positive  product. 
If  the  number  of  negative  factors  is  odd,  there  will  be  just  one  remain- 
ing, which  makes  the  final  product  negative. 

EXERCISES 

Find  the  products  of  the  following : 

1.  2« .  3* .  4:\  2''S''  42.  8.  a%  a^-^,  a^-^. 

2.  3  .  2^ .  52,  5 .  22 . 5,  7  .  2^ .  53.    9.  a"6-,  a^h^,  a'-^'^b'-^. 

3.  2  spYj  5  s^f,  2  x*y.  10.  4  a  6«  2  a^b%  3  a^ft^-m-n 

4.  5  an/,  2  x^y,  4  xf,  a^y\  11.  2  0;'"?/"'+",  3  x"''y''-"'+\ 

5.  3  a'bc,  ab%  a^bc\  4  ab^c.     12.  a'^-^^%rn-3n^  ^2c-d-ij^2-n.+sn^ 

6.  a;",  a;-i,  a;"+^  2  af .  13.  3  a^+^  2  x^'^f^  2  x^-^-^f-^\ 

7.  a^+"-^  aj'^-^+i,  x'^\  14.  a^-^l^'+\  a'+^¥-\  Sa^b\ 

15      3*«-2— 2t  .  2»+3-m    05— 4a+26  .  9«i+2-n 

16.  a^+i+y,  sH'-^-^y^,  y^-^,         17.  7  •  2**-*,  3  •  2«-2«,  5  •  23-«. 
18.  2^— ^-%  3  .  21-**-^  32 .  22-2*. 

20.  (1  -f-  a)7-*+«  •  (1  -  a)2+«-'',  (1  -  o)"-"-^  •  (1  -h  a)»— «. 


296  FUNDAMENTAL  OPEKATIONS 


DIVISION   OF   MONOMIALS 

46.  In  the  Elementary  Course  we  have  seen  that  a;^  -f-  a?^  = 
a;®-*  =  x^,  etc.  In  general,  if  a  is  any  number  and  m  and  k  are 
any  positive  integers,  of  which  m  is  the  greater,  then 

fl/M  _j_  flA  =  a'"~*. 

For,  since  k  and  m  —  k  are  both  positive  integers,  we  have,  by  §  43, 
Qi^m-*  —  (jik+m-k  —  qto^  That  is,  a"*-*  is  the  number  which  multiplied 
by  a*  gives  a  product  a"»,  and  hence  by  the  definition  of  division, 

Qfn,  _^  (jk  —  (fii—k^ 

0 

Hence :  The  quotient  of  two  powers  of  the  same  base  is  a  power 
of  that  base  whose  exponent  is  the  exponent  of  the  dividend  minus 
that  of  the  divisor. 

Under  the  proper  interpretation  of  negative  numbers  used 
as  exponents  this  principle  also  holds  when  m  <  Jc.  This  is 
considered  in  detail  in  §  177.  We  remark  here  that  in  case 
m  =  k,  the  dividend  and  the  divisor  are  equal  and  the  quotient 


is  unity.     Hence  a"'-i-a'^  =  a"'"'"  =  a^ 

=  1.     See  §11. 

47.  We  have  seen  in  the  earlier  work  that  —  =  -^,  etc. 

^y    ^y 

In  general,  if  a,  6,  and  k  are  any  number  expressions : 

ak     a 
bk     b' 

' 

For,  by  definition  of  division, 

«=5.5.            (1) 

Multiplying  both  sides  of  (1)  by  ^, 

ak  =  -'  bk.                       (2) 
b 

Dividing  both  sides  of  (2)  by  bk, 

«^  =  «.                            (3) 
bk      b                              ^  ^ 

Hence:  In  dividing  one  algebraic  expression  by  another,  all 
factors  common  to  dividend  and  divisor  may  be  removed  or 
canceled. 


MULTIPLICATION  AND   DIVISION  297 

Tx.    -1  EXERCISES 

Divide : 

1.  4.2^.3^.52  by  3. 28. 3*. 5.  4.  5a«6V  by  5a*b'^c*c?, 

2.  6. 3^7^.  13*  by  2. 3*.  72. 132.  5.  a^»y"*z^  by  x^y'^TT, 

3.  Sx'yh  by  2a^yz.  6.  a^-'y^+^  by  a^+'y^+y 

g      ^0+26-7  .  ^35-2o+4    jrjy    ^i+a-8  .  g26-2a+3^ 
9.     a^+2«-3n55gr-n    ]^y    ^2+m-4n54g7-n^ 

11    2^"^"^^*  •  3*-<<'+^  bv  2^"*"'*  •  3*-*^+^. 

12.  (X  —  2)3'"+^-3«  .  (x  +  2)^'"+2-3n    by   (^a;  _|.  2)l+3»-2n  .  (g.  _  2)l-3"+2«^ 

13.  (x-y)""-^-^ .  (a; +  2//'=-''"''  by  (x-y)-^-^^''  -  (x -{- y)-^'^-^K 

14.  (a*  -  62)8+4fc+76  .  (^2  _  yiy-zk-sb  ^y  (a?^}^u-k .  (^2  _  52^-2+2*^ 

MULTIPLICATION   OF  POLYNOMIALS 

48.  In  §  87,  E.G.,  we  saw  that  the  product  of  two  polynomials 
is  equal  to  the  sum  of  the  products  obtained  by  multiplying  each 
term  of  one  polynomial  by  every  term  of  the  other. 

This  follows  from  the  distributive  law  of  multiplication,  §  10.  For 
by  this  law, 

Qn-{-  n  -\-  k)(a  +  &  +  c)  =  m(a  +  6  +  c)     •^ 

Applying  the  same  law  to  each  part,  we  have  the  product, 
ma  +  mb  -^  mc  -\-  na  ->(■  nh  ■\-  nc  •\-  ka  ■\-  Tcb  -^  kc. 


298  FUNDAMENTAL   OPERATIONS 

EXERCISES 

Find  the  following  indicated  products : 

1.  (a  +  &)  («  +  b),  i.e.  (a  +  bf ;  also  (a  -  bf. 

2.  (a  +  6)  (a  +  b)  (a  +  5),  i.e.  (a  -f  6)^ ;  also  (a  -  bf. 

3.  (a  +  6)  (a  +  5)  (a  +  5)  (a  +  6),  i.e.  (a  +  6)^ ;  also  (a  —  b)\ 

4.  (a-  +  2  a6  +  6-)  (a'  +  2  a6  +  b^)  (a  +  6). 

5.  {a'  -2ab  +  b^(a'-2ab-\-  b')  (a  -  b). 

Note  that  Ex.  4  gives  (a  -\-  by  and  Ex.  5  gives  (a  —  6)*. 

6.  (a^  +  2  a6  +  6^)3 ;  also  (a  +  6)1 

7.  (a^  -  2  a6  +  fe^^^ .  ^Iso  (a  -  bf. 

8.  Collect  in  a  table  the  following  products  : 

(a-^bf,  (a -by,  (a  +  by,  (a -by,  (a -{-by. 

(a -by,        (a  +  by,        (a -by,        (a  +  by,        (a -by. 

9.  From  the  above  table  answer  the  following  questions : 

(a)  How  many  terms  in  each  product,  compared  with  the 
exponent  of  the  binomial  ? 

(b)  Tell  how  the  signs  occur  in  the  various  cases. 

(c)  How  do  the  exponents  of  a  proceed  ?  of  6  ? 

(d)  Make  a  table  of  the  coefficients  alone  and  memorize  this. 

E.g.   For  (a  +  by,  they  are  1,  5,  10,  10,  5,  1. 

10.  From  the  above  question  make  rules  for  finding  the 
powers  of  a  binomial  up  to  the  fifth.  Apply  these  rules  to  the 
following  examples  : 

(a)  (2  a;  +  3  yf.  (/)  (^  a«  - 1  byf.       ,      (a  _  J' 

(b)  (3  s  -  5  yf.  (g)  (2  to  -  nf.  '     \2 

^"'[z    sj-  (''>  [s-l)-  (r)(i-2xy. 

(d)(2x  +  3yy.  (         yy_  (m)(l+3xy. 

(,e)(3a-by.  «i2^-2J 

11.  {a  +  b  +  cy.  12.    (a  +  b~cy.  13.   (a-b-cy 


MULTIPLICATION  AND  DIVISION  299 

14.  From  Exs.  11-13  deduce  a  rule  for  squaring  a  trinomial, 
and  apply  it  to  the  following : 

(a)    (2x-3y-^4.zy,  (6)    {^m-^n  +  iry. 

15.  (x -\- y  -\-  z  -{-  vf.         16.    {x  —  y  +  z  —  vy, 

17.  From  Exs.  15,  16  deduce  a  rule  for  squaring  a  poly- 
nomial, and  apply  it  to  the  following  : 

(a)  (2x  +  y-\-z  +  SiL^f. 

(b)  (a^-{-Sa^b  +  3ab^-\-by. 

(c)  (a^-Sa'b-\-3ab^-by. 

18.  (a  -  6)  (a2 -f  a6 -h  52). 

19.  (a  +  b){a'-ab  +  b'-y 

20.  (a  -  b)  (a^  +  a'b  +  ab'  +  b^. 

21.  (a-\-b){a^-a^b-\-ab^-b'). 

22.  (a  -  b)  (a' -\- a^b -{- a^b' +  ab^  +  b*). 

23.  (a  +  b)  (a^  -  a%  +  aV/  -  ab^  +  6*). 

24.  (a  _  6)  (rt^  _j_  «4^  4.  ^3^2  _,.  ^253  ^  ^4  _^  55)^ 

25.  (a  +  b)  (a'  -  a'b  +  a^b^  -  a^b^  4-  «&*  -  b'). 

26.  (l-r)(a  +  a?-H-a?-^-|-a?-3). 

27.  (l  —  r){a  +  ar-{-ar^-\-ai^-\-ar^-\-aT^). 

DIVISION  OP  POLYNOMIALS 

49.  According  to  the  distributive  law  of  division,  §  30,  a 
polynomial  is  divided  by  a  monomial  by  dividing  each  term 
separately  by  the  monomial. 

■^  ^  ah  •}■  ac  —  ad     ah  ,  ac     ad     ,   ,         j 

J^ .g.  — 1 =o  +  c  —  rf. 

a  a       a       a 

A  polynomial  is  divided  by  a  polynomial  by  separating  the 
dividend  into  polynomials,  each  of  which  is  the  product  of  the 
divisor  and  a  monomial.  Each  of  these  monomial  factors  is  a 
.part  of  the  quotient,  their  sum  constituting  the  whole  quotient. 
The  parts  of  the  dividend  are  found  one  by  one  as  the  work  pro- 
ceeds.   See  §§  148-149,  E.G.    This  is  best  shown  by  an  example. 


300  FUNDAMENTAL  OPERATIONS 

a2  +  2a-3,  Divisor. 


a^  —  a  +  1,     Quotieut. 


Dividend,  a*  +     aS  -  4  a^  +  5  a  -  3 

1st  part  of  dividend  :   a^  +  2a^-Sa'^ 

-a^-a^    -f5a-3 
2d  part  of  dividend :  —  a^  —  2  g^  +  3  a 

a2  +  2  a  -  3 
3d  part  of  dividend :  a^  +  2  a  -  3 

0 

The  three  parts  of  the  dividend  are  the  products  of  the 
divisor  and  the  three  terms  of  the  quotient.  If  after  the  suc- 
cessive subtraction  of  these  parts  of  the  dividend  the  remain- 
der is  zero,  the  division  is  exact.  In  case  the  division  is  not 
exact,  there  is  a  final  remainder  such  that 

Dividend  =  Quotient  x  divisor  +  Remainder, 

In  symbols  we  have  D  =  Q  •  d  -|-  R. 

Divide:  EXERCISES 

1.  0^  +  5  a;*?/  +  10  x^y^  + 10  a^f  -\-5xy^  -\-  f  by  x^  +2  xy-\-f, 

2.  y^-\-x'^y^-\-y^hj  x'^—x^'if-\-y^.   6.  oc^  —  y^  by  a^  —  y\ 

3.  xP-f  by  x-y.  7.  a^  +  b^  by  a^-\-b\ 

4.  x^-f  by  a^-\-a^y-\-xy^+f.     8.  aj««  -  y^  by  a^"  -  y^. 

5.  a^^y^  hj  x^-xy-{-  yK  9.  a}^-a'b'+b^'' by  a'-ab-^-b^ 

10.  2a^-3ar'6-f  6a;252_a;63  +  66*  by  a^ - 2 a;6 -|- 3 6^ 

11.  2  0^-5x^^+6  x*-6x^-{-6x'^-4.x-{-l  by  a^^-aj^+aj^-aj+l. 

12.  26  a^68  +  a«  +  6  6«  -  5  a«6  -  17  a6^  -  2  a^^^  _  ^254 

by  a'-Sb^-2ah 

13.  x'  +  2x^-7x'~-Sx-\-12  hy  ay'-Sx-\-2. 

14.  4  62-f  4 a6  +  a2- 12 5c- 6 ac-f  9 c2  by  2  6 -t- a -3c. 

15.  aj^  +  4  a?/^  —  4  xyz  +  3  y*  +  2  yh  —  z^  hj  a^  —  2  xy  +  3  y^—z. 

16.  a262c  +  3  a^b^  -3abc^-  a?&  +  6'  -  4  b^'c^  -f  3  aft^c 

-f36c*-3a26c2  by  d^^c^. 


CHAPTER   III 

INTEGRAL  EQUATIONS  OF  THE  FIRST   DEGREE  IN  ONE 
UNKNOWN 

50.  When  in  an  algebraic  expression  a  letter  is  replaced  by 
another  number  symbol,  this  is  called  a  substitution  on  that  letter. 

E.g.  In  the  expression,  2  a  +  5,  if  a  is  replaced  by  3,  giving  2-3  +  5, 
this  is  a  substitution  on  the  letter  a. 

51.  An  equality  containing  a  single  letter  is  said  to  be 
satisfied  by  any  substitution  on  that  letter  which  reduces  both 
members  of  the  equality  to  the  same  number. 

E.g.    4  re  +  8  =  24  is  satisfied  by  a:  =  4,  since  4  •  4  +  8  =  24. 

We  notice,  however,  that  the  substitution  must  not  reduce  the 
denominator  of  any  fraction  to  zero. 

Thus  X  z=2  does  not  satisfy —  =  8  although  it  reduces  the  left 

0  ^  ~ . 
member  of  the  equation  to  -,  which  by  §  24  equals  8  or  any  other 

number  whatever. 

On  the  other  hand,  a;  =  6  satisfies  this  equation,  since 

62-4     32 


6-2  ~  4 


8. 


52.  An  equality  in  two  or  more  letters  is  satisfied  by  any 
simultaneous  substitutions  on  these  letters  which  reduce  both 
members  to  the  same  number. 

E.g.  6  a  +  3  &  =  15  is  satisfied  bya  =  2,  &  =  l;a  =  |,  &  =  2;a  =  l, 
6  =  3,  etc. 

x'^-y''       _  1  : 


„      ^  „     ^  is  satisfied  by  a:  =  3,  y  =  1,  but  is  not  satisfied  by 

x^  -{■  2  xy  +  y^     2 

any  values  of  x  and  y  such  that  x  =  —  y,  since  these  reduce  the 

denominator  (and  also  the  numerator)  to  zero.     See  §  24. 

301 


302      INTEGRAL   EQUATIONS   OF  THE   FIRST   DEGREE 

53.  An  equality  is  said  to  be  an  identity  in  all  its  letters,  or 
simply  an  identity,  if  it  is  satisfied  by  every  possible  substitiir 
tion  on  these  letters,  not  counting  those  which  make  any 
denominator  zero. 

If  an  equality  is  an  identity,  both  members  will  be  reduced 
to  the  same  expression  when  all  indicated  operations  are  per- 
formed as  far  as  possible. 

The  members  of  an  identity  are  called  identical  expressions. 

Thus  in  the  identity  (a  +  hY=a'^-\-2  ab  +  h'^,  performing  the  indi- 
cated operation  in  the  first  member  reduces  it  to  the  same  form  as  the 
second. 

54.  An  equality  which  is  not  an  identity  is  called  an  equa- 
tion of  condition  or  simply  an  equation. 

The  members  of  an  equation  cannot  be  reduced  to  the  same 
expression  by  performing  the  indicated  operations. 

E.g.  (a;  —  2)  (a;  —  3)  =  0  cannot  be  so  reduced.  This  is  an  equation 
which  is  satisfied  by  a;  =  2  and  a:  =  3.     See  §  22. 

55.  In  an  equation  containing  several  letters  any  one  or  more 
of  them  may  be  regarded  as  unknown,  the  remaining  ones  being 
considered  known.  Such  an  equation  is  said  to  be  satisfied  by 
ajiy  substitution  on  the  unknowyi  letters  which  reduces  it  to  an 
identity  in  the  remaining  letters. 

E.g.  x^  —  t^  =  sx  +  St  is  an  equation  in  s,  x,  or  t,  or  in  any  pair  of 
these  letters,  or  in  all  three  of  them. 

As  an  equation  in  x  it  is  satisfied  by  a:  =  s  +  ;,  since  this  substitu- 
tion reduces  it  to  the  identity  in  s  and  t, 

s^  +  2  st  =  s^  +  2  St.  ^ 

As  an  equation  in  s  it  is  satisfied  hj  s  =  x  —  i,  since  this  substitu- 
tion reduces  it  to  the  identity  in  x  and  t, 

a:2- <2  =  a^2  _  ^2. 

Any  number  expression  which  satisfies  an  equation  in  one 
unknown  is  called  a  root  of  the  equation. 

E.g.  s  +  /  is  a  root  of  the  equation  x^  —  t^  =  sx  +  st,  when  x  is  the 
t  is  a  root  when  s  is  the  unknown. 


EQUIVALENT  EQUATIONS  303 

56.  An  equation  is  rational  in  a  given  letter  if  every  term  in 
the  equation  is  rational  with  respect  to  that  letter. 

An  equation  is  integral  in  a  given  letter  if  every  term  is 
rational  and  integral  in  that  letter. 

57.  The  degree  of  a  rational,  integral  equation  in  a  given 
letter  is  the  highest  exponent  of  that  letter  in  the  equation. 

In  determining  the  degree  of  an  equation  according  to  this  defini- 
tion it  is  necessary  that  all  indicated  multiplications  be  performed  as 
far  as  possible. 

E.g.  (a:  —  2)  (x  —  3)  =  0  is  of  the  2d  degree  in  x,  since  it  reduces  to 
a:2-5a;  +  6  =  0. 

EQUIVALENT   EQUATIONS 

58.  Two  equations  are  said  to  be  equivalent  if  every  root  of 
either  is  also  a  root  of  the  other. 

In  the  Elementary  Course,  §  36,  we  found  that  an  equation 
may  be  changed  into  an  equivalent  equation  by  certain  opera- 
tions, which  are  now  further  considered  in  principles  1,  2,  and 
3  below : 

59.  Principle  1.  If  one  rabional,  integral  equation  is 
derived  from  anotJier  hij  performing  the  indieated  operor 
tions,  then  the  two  equations  are  equivalent. 

This  is  evident,  since  in  performing  the  indicated  operations  each 
expression  is  replaced  by  another  identically  equal  to  it.  Hence  any 
expression  which  satisfies  the  given  equation  must  satisfy  the  other, 
and  conversely. 

E.g.  10  a;  =  50  is  equivalent  to  3  ar  +  7  a:  =  50,  since  3  a;  +  7  a;  =  10  a: ; 
and  8(2  a;-3y)  =  2y  —  lis  equivalent  to  16  a:  —  24  y  =  2  y  —  1,  since 
8(2a:-3y)  =  16a:-24y. 

60.  Principle  2.  If  any  equation  is  derived  from,  an- 
other by  adding  the  same  expression  to  each  member,  or 
by  subtracting  the  same  expression  from  each  member, 
then  the  equations  are  equivalent. 

For  simplicity  consider  an  equation, 

M=N,  (1) 


304      INTEGRAL  EQUATIONS   OF  THE   FIRST  DEGREE 

containing  only  one  unknown,  x.  Add  to  each  member  an  expression 
i4,  which  may  or  may  not  contain  x. 

Then  M  +  A=N-^-A  (2) 

is  easily  seen  to  have  the  same  roots  as  (1).  For  if  a  certain  value  of 
X  makes  M  equal  N,  this  will  also  make  M  ■\-  A  equal  N  -^  A,  since 
any  value  whatever  of  x  makes  A  equal  A.  Hence  any  number  which 
is  a  root  of  (1)  is  also  a  root  of  (2). 

Again,  any  value  of  x  which  makes  M  +  A  =  N  -\-  A  will  also 
make  M  equal  N,  since  every  value  of  x  makes  A  equal  A.  Hence 
any  number  which  is  a  root  of  (2)  is  also  a  root  of  (1).  We  have 
there f© re  shown  that  (1)  and  (2)  are  equivalent  according  to  the  defi- 
nition, §  58. 

This  argument  is  based  on  axioms  I  and  IV.  By  use  of  the  same 
axioms  we  may  show  that  M  —  A  =  N  —  A  and  M  =  N  are  equiva- 
lent equations. 

61.  It  follows  that  any  equation  can  be  reduced  to  an  equiva' 
lent  equation  of  the  form  R  =  0. 

For  if  an  equation  is  in  the  form  M  =  N,  then  by  principle  2  it  is 
equivalent  to  M-N=N  —  N=0,  which  is  in  the  form  R  =0. 

62.  Principle  3.  If  one  equation  is  derived  from  another 
hy  multiplying  or  dividing  each  member  by  the  same  ex- 
pression, then  the  equations  are  equivalent,  provided  the 
original  equation  is  not  multiplied  or  divided  hy  zero  or 
hy  an  expression  containing  any  of  the  unhnowns  of  the 
equation. 

This  principle  follows  from  axioms  V  and  IX  by  argument  similar 
to  that  used  in  §  60.  In  this  case,  however,  the  expression  A  must 
not  contain  x  and  must  not  be  zero,  as  was  possible  in  principle  2. 

E.g.  a:  +  1  =  5  and  (jjc  —  l)(a;  +  1)  =  5(a:  —  1)  are  not  equivalent 
equations,  since  the  first  has  only  the  root  a;  =  4,  while  the  second 
has  in  addition  the  root  a;  =  1,  as  may  be  easily  verified. 

Similarly,  x  —  o  =  l  and  0  •  (z  —  5)  =  0.7  are  not  equivalent,  since 
the  first  has  only  the  root  x  =  12,  while  the  second  is  satisfied  by  any 
number  whatever. 


EQUIVALENT  EQUATIONS  305 

63.  The  ordinary  processes  of  solving  equations  depend 
upon  principles  1,  2,  and  3,  as  is  illustrated  by  the  following 
examples : 

Ex.1.  (a:+4)(x  +  5)  =  (x+2)(x  +  6).  (1) 

a:2  +  9  X  +  20  =  x2  +  8  a:  +  12.  (2) 

x  =  -8.  (3) 

By  principle  1,  (1)  and  (2)  are  equivalent,  and  by  principle  2,  (2) 
and  (3)  are  equivalent.     Hence  (1)  and  (3)  are  equivalent.     That  is, 

—  8  is  the  solution  of  (1). 

Ex.2.                              ix  +  ^  =  i:  (1) 

22:  + 4  =  12.  (2) 

2a:  =  8.  .      (3) 

a:  =  4.  (4) 

By  principle  3,  (1)  and  (2)  are  equivalent.  By  principle  2,  (2)  and 
(3)  are  equivalent.  By  3,  (3)  and  (4)  are  equivalent.  Hence  (1) 
and  (4)  are  equivalent  and  4  is  the  solution  of  (1). 

These  principles  are  stated  for  equations,  but  they  apply 
equally  well  to  identities,  inasmuch  as  the  identities  are 
changed  into  other  identities  by  these  operations. 

64.  If  an  identity  is  reduced  to  the  form  72  =  0,  §  61,  and  all 
the  indicated  operations  are  performed,  then  it  becomes  0  =  0. 
See  §  53.  Conversely,  if  an  equality  may  be  reduced  to  the 
form  0  =  0,  it  is  an  identity.  This,  therefore,  is  a  test  as  to 
whether  an  equality  is  an  identity. 

E.g.  (a:  +  4)2  =  a:2  +  8x  +  16  is  an  identity,  since  in  a:^  +  8a:  +  16 

—  a:^  —  8  a:  —  16  =  0  all  terms  cancel,  leaving  0  =  0. 

EXERCISES 

In  the  following,  determine  which  numbers  or  sets  of  num- 
bers, if  any,  of  those  written  to  the  right,  satisfy  the  corre- 
sponding equation. 

Remember  that  no  substitution  is  legitimate  which  reduces 
any  denominator  to  zero. 


306      INTEGRAL   EQUATIONS   OF  THE   FIRST   DEGREE 

1.   A(x-l)(x-2)(x-S)=S(x-2Xx-3).  1,2,3,4 

'•  Mi— I-        ,  '''''■ 

6.  3a  +  46  =  12.  1'*  =  ^'    1"  =  ^'   1"  =  ^' 

[6  =  3.    |6  =  0.    [6=2. 

369(a-6)^        .  («  =  0,    |a  =  l,    |a  =  5, 

■       a^'  +  fe'^  ■  16  =  0.    [6=1.    [6  =  4. 

10.  <-;t7-!i:y =(»--3-)(2.-3.).  .=M=i; 

r  =  l,  s=— 1;     r=2,  s  =  — 2;     y  =  a,  s  =  — a. 

11.   a  +  6  +  c  =  6.     a  =  l,  6=2,  c  =  3; 

a  =  3,6  =  3,c  =  0;     a  =  10,  6  =  0,  c= -4. 

,o       o^-^-c     _3a-2c  +  5&  +  2       ^      o   .      r.  ^     « 
^^-    V^M^6^^" 10 a  =  8,6  =  0,c  =  6; 

a  =  l,  6  =  4,  c  =  2;     a  =  0,  6  =0,  c=-4. 

^3    (a-6)(6-c)(c-a)^ 

ac  —  6c  —  a^  +  6a  ^  ' 

a  =  3,  6  =  2,c  =  3;     a  =  6,  6  =  6,  c  =  0. 


9 


EQUIVALENT   EQUATIONS  307 

14.  {x-z){x-y){y-z)  =  %xyz{^-y^{f-z''){z'~a?). 
f 

x  =  lj  y  =  0,  z  =  l;  x  =  l,  y  =  2y  z=:S. 

15.  a^-^Sx^y  +  3xf  +  f={x-{-yy.    j"'"^'   1^  =  ^' 

16.  Show  by  reducing  the  equality  in  Ex.  15  to  the  form 
R  =  0  that  it  is  satisfied  by  any  pair  of  values  whatsoever  for 
X  and  y,  e.g.,  for  a;  =  348764,  2/  =  594021.  What  kind  of  an 
equality  is  this  ? 

Which  of  the  following  four  equalities  are  identities  ? 

17.  12(x-\-yy  +  17(x-{-y)-7  =  {Sx-\-3y-l)(4.x-\-4.y+7y 

18.  9^JZ^  =  a'-^a^b-\-a'b^-\-ab'-{-b\ 

a  —  b 

19.  t±J^=:a^-a'b-\-a'b'-ab^-{-b*. 
a-\-b 

20.  2(a-&)2  +  5(a  +  6)4-8a6  =  (2a  +  26  +  l)(tt  +  6  4-l). 
Solve  the  following  equations,  and  verify  results  in  21-25 : 

21.  (2a  +  3)(3a-2)  =  a2  +  a(5aH-3). 

22.  6(6-4)2=-5-(3-2  6)2-5(2  +  6)(7-2  6). 

23.  (y  -  Sy  +  (y  -  4)=^  -(y-  2y  -  (y  -  Sy  =  0. 

24.  (a;-3)(3a;-}-4)-(a;-4)(a;-2)  =  (2ic-|-l)(a;-6). 

25.  2(3r--2)(4r  +  l)-h(r-4)s  =  (rH-4)3-2. 

26.  a^—c-^b^c-{-  abc  =  b.     (Solve  for  c.) 

In  the  next  three  examples  solve  for  y. 

27.  (b-2y(b-y)-Sby -^  (2b -\-l)(b-l)  =  3 -2b. 

28.  ny  (y -{.  n)  -  (y -{-  m)  (y  +  n)  (m  +  n)  +  my  (y  -f  m)  =  0. 

29.  (m  +  n) (n -^  b  —  y) -\-  (n  —  m){b  —  y)=  n (m  +  6). 
In  the  next  eight  examples  solve  for  x. 


308      INTEGRAL   EQUATIONS   OF  THE  FIRST  DEGREE 

30.  2  (12  -  05)  +  3  (5  a;  -  4)  4-  2  (1 6  -  aj)  =  12  (S  +  x). 

31.  (b  -  a)x  —  (a -}-b)x -{- 4: or  =  0. 

32.  (x  —  a)(b  —  c)  +  (b  —  a)  {x  —  c)  —  {a  —  c)  (a; — 6)  =  0. 

33.  (a;-3)(a;-7)-(a;-5)(x-2)+12  =  2(a:-l). 

34.  (a  +  bf  +{x-b)(x-a)-(x-\- a) {x  +  5)  =  0. 

35.  37^(5a;-l)4-A(2-3a;)+i(4  +  a5)=t(H-2a;)-^. 

36.  a(x  —  b)  —  {a  +  b){x  +  b  —  a)-=b{x  —  a)+o?  —  W, 

37.  (I  —m)(x  —  n)  -\-2l {m -\- n)  =  (I  +  m) {x  +  n). 

Solve  each  of  the  following  equations  for  each  letter  in  terms  of 
the  others : 

38.  l{W+w')  =  VW\  40.    m2S2(^2-O  =  (^^  +  ^i)0-^i)- 

39.  (v  —  n)d  =  {v  —  n^di.     41.    (m  +  wii)  (^i  —  Q  =  Zm2  +  m^^. 

PROBLEMS 

1.  What  number  must  be  added  to  eacb  of  the  numbers  2, 
26,  10  in  order  that  the  product  of  the  first  two  sums  may- 
equal  the  square  of  the  last  sum  ? 

2.  What  number  must  be  subtracted  from  each  of  the 
numbers  9,  12,  18  in  order  that  the  product  of  the  first 
two  remainders  may  equal  the  square  of  the  last  re- 
mainder ? 

3.  What  number  must  be  added  to  each  of  the  numbers 
a,  6,  c  in  order  that  the  product  of  the  first  two  sums  may 
equal  the  square  of  the  last  ? 

Note  that  problem  1  is  a  special  case  of  3.     Explain  how  2  may 
also  be  made  a  special  case  of  3. 

4.  What  number  must  be  added  to  each  of  the  numbers 
a,  b,  c,  d  in  order  that  the  product  of  the  first  two  sums  may 
equal  the  product  of  the  last  two  ? 


PROBLEMS  IN  ONE  UNKNOWN  S09 

6.  State  and  solve  a  problem  which  is  a  special  case  of 
problem  4. 

6.  What  number  must  be  added  to  each  of  the  numbers 
a,  b,  c,  d  in  order  that  the  sum  of  the  squares  of  the  first  two 
sums  may  equal  the  sum  of  the  squares  of  the  last  two  ? 

7.  State  and  solve  a  problem  which  is  a  special  case  of 
problem  6. 

8.  What  number  must  be  added  to  each  of  the  numbers 
a,  b,  c,  d  in  order  that  the  sum  of  the  squares  of  the  first  two 
sums  may  be  k  more  than  twice  the  product  of  the  last  two  ? 

9.  State  and  solve  a  problem  which  is  a  special  case  of 
problem  8. 

10.  The  radius  of  a  circle  is  increased  by  3  feet,  thereby  in- 
creasing the  area  of  the  circle  by  60  square  feet.  Find  the 
radius  of  the.  original  circle. 

The  area  of  a  circle  is  irr^.    Use  3j  for  tt. 

11.  The  radius  of  a  circle  is  decreased  by  2  feet,  thereby 
decreasing  the  area  by  36  square  feet.  Fin^  the  radius  of  the 
original  circle. 

12.  State  and  solve  a  general  problem  of  which  10  is  a 
special  case. 

13.  State  and  solve  a  general  problem  of  which  11  is  a 
special  case. 

How  may  the  problem  stated  under  12  be  interpreted  so  as  to 
include  the  one  given  under  13? 

14.  Each  side  of  a  square  is  increased  by  a  feet,  thereby 
increasing  its  area  by  b  square  feet.  Find  the  side  of  the 
original  square. 

Interpret  this  problem  if  a  and  b  are  both  negative  numbers. 

15.  State  and  solve  a  problem  which  is  a  special  case  of  14, 
(1)  when  a  and  b  are  both  positive,  (2)  when  a  and  b  are  both 
negative. 


310      INTEGRAL   EQUATIONS   OF  THE   FIRST  DEGREE 

16.  Two  opposite  sides  of  a  square  are  each  increased  by  a 
feet  and  the  other  two  by  h  feet,  thereby  producing  a  rectangle 
whose  area  is  c  square  feet  greater  than  that  of  the  square. 
Find  the  side  of  the  square. 

Interpret  this  problem  when  a,  h,  and  c  are  all  negative  numbers. 

17.  State  and  solve  a  problem  which  is  a  special  case  of  16, 
(1)  when  a,  6,  and  c  are  all  positive,  (2)  when  a,  h,  and  c  are 
all  negative. 

18.  A  messenger  starts  for  a  distant  point  at  4  a.m.,  going 
5  miles  per  hour.  Four  hours  later  another  starts  from  the 
same  place,  going  in  the  same  direction  at  the  rate  of  9  miles 
per  hour.  When  will  they  be  together  ?  When  will  they  be 
8  miles  apart  ?     How  far  apart  will  they  be  at  2  p.m.  ? 

For  a  general  explanation  of  problems  on  motion,  see  p.  95,  E.  C. 

19.  One  object  moves  with  a  velocity  of  v^  feet  per  second 
and  another  along  the  same  path  in  the  same  direction  with  a 
velocity  of  V2  feet.  If  they  start  together,  how  long  will  it  re- 
quire the  latter  to  gain  n  feet  on  the  former  ? 

From  formula  (2\  p.  96,  E.  C,  we  have  t  =  — 

Discussion.  If  V2  >  h\  and  n  >  0,  the  value  of  t  is  positive,  i.e.  the 
objects  will  be  in  the  required  position  some  time  after  the  time  of 
starting. 

If  V2  <  yi  and  n  >  0,  the  value  of  t  is  negative,  which  may  be  taken 
to  mean  that  if  the  objects  had  been  moving  before  the  instant  taken 
in  the  problem  as  the  time  of  starting,  then  they  would  have  been  in 
the  required  position  some  time  earlier. 

If  i?2  =  vi  and  n=^0,  the  solution  is  impossible.  See  §  25.  This 
means  that  the  objects  will  never  be  in  the  required  position.  If 
vi  =  V2  and  n  =  0,  the  solution  is  indeterminate.  See  §  24.  This  may 
be  interpreted  to  mean  that  the  objects  are  always  in  the  required 
position. 

20.  State  and  solve  a  problem  which  is  a  special  case  of  19 
under  each  of  the  conditions  mentioned  in  the  discussion. 

21.  At  what  time  after  5  o'clock  are  the  hands  of  a  clock 
first  in  a  straight  line  ? 


PROBLEMS  IN  ONE  UNKNOWN  311 

22.  Saturn  completes  its  journey  about  the  sun  in  29  years 
and  Uranus  in  84  years.  How  many  years  elapse  from  con- 
junction to  conjunction?   ^See  figure,  p.  269,  E.  C. 

23.  An  object  moves  in  a  fixed  path  at  the  rate  of  v^  feet 
per  second,  and  another  which  starts  a  seconds  later  moves  in 
the  same  path  at  the  rate  of  Vo  feet  per  second.  In  how  many 
seconds  will  the  latter  overtake  the  former  ? 

24.  In  problem  23  how  long  before  they  will  be  d  feet 
apart  ? 

If  in  problem  24  d  is  zero,  this  problem  is  the  same  as  23*  If  d  is 
not  zero  and  a  is  zero,  it  is  the  same  as  problem  19. 

25.  A  beam  carries  3  weights,  one  at  each  end  weighing 
100  and  120  pounds  respectively,  and  the  third  weighing  150 
pounds  2  feet  from  its  center,  where  the  fulcrum  is.  What  is 
the  length  of  the  beam  if  this  arrangement  makes  it  balance  ? 

For  a  general  explanation  of  problems  involving  the  lever,  see  pp. 
98  and  270,  E.  C. 

26.  A  beam  whose  fulcrum  is  at  its  center  is  made  to  bal- 
ance when  weights  of  60  and  80  pounds  are  placed  at  one  end 
and  2  feet  from  that  end  respectively,  and  weights  of  50  and 
100  pounds  are  placed  at  the  other  end  and  3  feet  from  it 
respectively.     Find  the  length  of  the  beam. 

27.  A  man  weighing  190  pounds  is  trying  to  pry  up  a  rock 
by  use  of  a  plank  12  feet  long  and  a  block  on  which  the 
plank  rests  as  a  fulcrum.  Find  the  weight  of  the  stone  if  he 
can  just  lift  it  when  the  fulcrum  is  2  feet  from  the  stone. 
How  heavy  a  stone  could  he  lift  by  putting  the  fulcrum  3 
feet  away  from  it  ?     Four  feet  away  from  it  ? 

28.  A  man  can  do  a  piece  of  work  in  16  days,  another  in  18 
days,  and  a  third  in  15  days.  How  many  days  will  it  require 
all  to  do  it  when  working  together  ? 

29.  A  can  do  a  piece  of  work  in  a  days,  B  can  do  it  in  b 
days,  C  in  c  days,  and  D  in  d  days.  How  long  will  it  require 
all  to  do  it  when  working  together  ? 


312      INTEGRAL  EQUATIONS  OF  THE   FIRST  DEGREE 


PROBLEMS  ON  THERMOMETER  READINGS 

There   are    two   kinds    of    thermometers    in   use    in    this 
country,  called  the  Fahrenheit  and  Centigrade,  the  former  for 


C    f 


ro 

60' 
60' 
•40' 
80° 
20' 
lO' 

o' 

-10 

-17.78' 


212* 
194° 
176* 
158' 
140' 
122* 
104* 


50 


common  purposes,  and  the  latter  for  scientific 
records  and  investigations.  Hence  it  frequently 
becomes  necessary  to  translate  readings  from 
one  kind  to  the  other. 

The  freezing  and  boiling  points  are  two  fixed 
temperatures  by  means  of  which  the  compu- 
tations are  made.  On  the  Centigrade  these  are 
marked  0°  and  100°  respectively  and  on  the 
Fahrenheit  they  are  marked  32°  and  212°  re- 
spectively. See  the  cut.  Hence  between  the 
two  fixed  points  there  are  100  degrees  Centi- 
grade and  180  degrees  Fahrenheit, 

That  is,  100  degree  spaces  on  the  Centigrade  cor- 
respond to  180  degree  spaces  on  the  Fahrenheit. 

Hence  1°  Centigrade  corresponds  to  f°  Fahrenheit^ 
or  P  Fahrenheit  corresponds  to  f°  Centigrade. 


All  problems  comparing  the  two  thermometers  are  solved  by 
reference  to  these  fundamental  relations. 

1.  If  the   temperature   falls   15   degrees   Centigrade,   how 
many  degrees  Fahrenheit  does  it  fall  ? 

2.  If  the   temperature   rises   18  degrees   Fahrenheit,  how 
many  degrees  Centigrade  does  it  rise? 

3.  Translate  -f-  25°  Centigrade  into  Fahrenheit  reading. 

25°  Centigrade  equals  §  •  25''  =  45°  Fahrenheit. 
45°  above  the  freeezing  point  =  45°  -f  32°  above  0°  Fahrenheit. 
Hence,  calling  the  Fahrenheit  reading  F,  we  have  F  =  32  +  f  •  25. 

4.  Translate  +14°  Centigrade  into  Fahrenheit  reading. 
Beasoning  as  before,  F  =  32  -f-  f  •  14. 


PROBLEMS  IN  ONE   UNKNOWN  313 

From  the  two  preceding  problems  we  have  the  formula 

F  =  32  +  |C.  (1) 

Translate  this  into  words,  understanding  that  F  and  C  stand 
for  readings  on  the  respective  thermometers. 

5.  Solve  the  above  equation  for  C  in  terms  of  F  and  find 

C  =  |(F-32).  (2) 

6.  Graph  the  equation  F  =  32-f  |C,  marking  Centigrade 
readings  on  the  horizontal  axis  and  Fahrenheit  on  the  vertical. 

From  the  graph  answer  the  following  questions  : 

7.  Find  the  Fahrenheit  reading  when  the  Centigrade  is 
zero;  10°;  20°;  30°;   -5°;   -15°. 

8.  Find  Centigrade  reading  when  the  Fahrenheit  is  32°; 
50°;  68°;  86°;  41°;  59°;  14°;  0°. 

It  thus  appears  that  the  graph,  so  far  as  it  extends,  shows  to 
the  eye  all  the  information  contained  in  the  equation 

F  =  32-ffC. 

9.  By  substituting  in  the  equation  F  =  32  +  |  C,  find  the 
Fahrenheit  reading  when  the  Centigrade  is  60°;  100°;  —40° 
or  the  freezing  point  of  mercury. 

10.  In  like  manner  find  the  Centigrade  reading  when  the 
Fahrenheit  is  -40°;  -20°;  98°  or  blood  heat;  212°  or  the 
boiling  point. 

11.  What  is  the  temperature  Centigrade  when  the  sum  of 
the  Centigrade  and  Fahrenheit  readings  is  102°  ? 

12.  What  is  the  temperature  Fahrenheit  when  the  sum  of 
the  Centigrade  and  Fahrenheit  readings  is  zero  ? 

13.  What  is  the  temperature  Centigrade  when  the  sum  of 
the  Centigrade  and  Fahrenheit  readings  is  140°  ? 

14.  What  is  the  temperature  in  each  reading  when  the 
Fahrenheit  is  50°  higher  than  the  Centigrade  ? 


CHAPTER  TV 

INTEGRAL  LINEAR  EQUATIONS  IN  TWO  OR  MORE 
UNKNOWNS 

INDETERMINATE   EQUATIONS 

65.  If  a  single  equation  contains  two  unknowns,  an  unlimited 
number  of  pairs  of  numbers  may  be  found  which  satisfy  the 
equation. 

E.g.  In  the  equation,  y  =  2  x  +  1,  by  assigning  any  value  to  a:,  a 
corresponding  value  of  y  may  be  found  such  that  the  two  together 
satisfy  the  equation. 

Thus,  a;=—  3,  y=  —  b;  xz=iO,  y=\'^x  =  2,  y  =  b,  are  pairs  of 
numbers  which  satisfy  this  equation. 

For  this  reason  a  single  equation  in  two  unknowns  is  called 
an  indeterminate  equation,  and  the  unknowns  are  called  varia- 
bles. A  solution  of  such  an  equation  is  any  pair  of  numbers 
which  satisfy  it. 

A  picture  or  map  of  the  real  (see  §§  135,  195)  solutions  of 
an  indeterminate  equation  in  two  variables  may  be  made  by 
means  of  the  graph  as  explained  in  §§  111,  112,  E.  G. 

66.  The  degree  of  an  equation  in  two  or  more  letters  is  the 
sum  of  the  exponents  of  those  letters  in  that  one  of  its  terms 
in  which  this  sum  is  greatest.     See  §  114,  E.  C. 

E.g.  y  =  2  a;  +  1  is  of  the^rs^  degree  in  x  and  y.  y^  =:2x  -\-  y  and 
y  =  2  xy  -\-  Z  are  each  of  the  second  degree  in  x  and  y. 

An  equation  of  the  first  degree  in  two  variables  is  called  a 
linear  equation,  since  it  can  be  shown  that  the  graph  of  every 
such  equation  is  a  straight  line. 

314 


INDETERMINATE   EQUATIONS  315 

67.  It  is  often  important  to  determine  those  solutions  of  an 
indeterminate  equation  which  are  positive  integers,  and  for  this 
purpose  the  graph  is  especially  useful. 

Ex.  1.     Find  the  positive  integral  solutions  of  the  equation 

Solution.  Graph  the  equation  carefully  on  cross-ruled  paper,  find- 
ing it  to  cut  the  ar-axis  at  x  =  14  and  the  ^-axis  at  y  =  6. 

Look  now  for  the  corner  points  of  the  unit  squares  through  which 
this  straight  line  passes.  The  coordinates  of  these  points,  if  there  are 
such  points,  are  the  solutions  required.  In  this  case  the  line  passes 
through  only  one  such  point,  namely  the  point  (7,  3).  Hence  the 
solution  sought  is  x  =  7,  y  =  d. 

Ex.  2.    Eind  the  least  positive  integers  which  satisfy 
7  x-Sy  =  17. 

Solution.  This  line  cuts  the  a:-axis  at  a:  =  2f  and  the  y-axis  at 
y  =  —  5|.  On  locating  these  points  as  accurately  as  possible,  the  line 
through  them  seems  to  cut  the  corner  points  (5,  6)  and  (8,  13).  The 
coordinates  of  both  these  points  satisfy  the  equation.  Hence  the 
solution  sought  is  x  =  5,  ^  =  6. 

EXERCISES 

Solve  in  positive  integers  by  means  of  graphs,  and  check : 

1.  x-\-y  =  7.  5.   90  — 5ic  =  9y. 

2.  x-^y  =  3.  6.   5x  =  29-Sy. 

3.  x-27=-9y.  7.   U0-7x-10y  =  0. 

4.  7!/- 112=  — 4a;.  8.   8  — 2a;-y  =  0. 

Solve  in  least  positive  integers,  and  check  : 
9.    7x  =  3y-^21.  11.    4:X  =  9y-36. 

10.   5a; -4?/ =  20.  12.    5x-2y-\-10  =  0. 

68.  In  the  case  of  two  indeterminate  equations,  each  of  the 
first  degree  in  Iwo  variables,  the  coordinates  of  the  point  of 
intersection  of  their  graphs  form  a  solution  of  both  equations. 


316  INTEGRAL  LINEAR  EQUATIONS 

Since  these  graphs  are  straight  lines,  they  have  only  one 
point  in  common,  and  hence  there  is  only  one  solution  of  the 
given  pair  of  equations. 

E.g.     On   graphing  x  +  y  =  4:  and  y  —  x  =  2,  the  lines  are  found 

to  intersect  in  the  point  (1,  3).     Hence  the  solution  of  this   ^.'\r  of 

equations  is  ^  o 

^  x=l,y  =  3. 

EXERCISES 

Graph  the  following  and  thus  find  the  solution  of  each  pair 
of  equations.     Check  by  substituting  in  1»he  equations. 

^^    \Sx-2y  =  -2,  ^^    fSx  =  7y, 


x-\-7y  =  30.  [x  +  3  =  5y-\-3. 


2. 


|aJ  +  2/  =  2,  g  \y  =  l, 

\3x-^2y  =  3.  '    '  [3y-^4.x  =  y. 

'    \2x  —  y  =  —  5.  '  \x  —  y  =  6y—3, 

4.    1^  =  -^'  10  1^  =  ^' 

\2x-3y  =  l,  \y  +  x  =  S. 

4a^  =  22/  +  6,  ^^  jy  =  -S: 


6. 


x-5==y-l.  \3x  +  2y  =  3. 

{x  =  y-5,  ^2     [a;  =  -2, 

l5?/  =  a7  +  9.  12/  =  5. 


SOLUTION  BY  ELIMINATION 

69.  The  solution  of  a  pair  of  equations  such  as  the  foregoing 
may  be  obtained  without  the  use  of  the  graph  by  the  process 
called  elimination.     See  pages  116-121,  E.  C. 

70.  Elimination  by  substitution  consists  in  expressing  one 
variable  in  terms  of  the  other  in  one  equation  and  substituting 
this  result  in  the  other  equation,  thus  obtaining  an  equation  in 
which  only  one  variable  appears.     See  §  122,  E.  C. 


SOLUTION  BY  ELIMINATION  317 

71.  Elimination  by  addition  or  subtraction  consists  in  making 
the  coefficients  of  one  variable  the  same  in  the  two  equations 
(§  62),  so  that  when  the  members  are  added  or  subtracted  this 
variable  will  not  appear  in  the  resulting  equation.  See  §  121, 
E.  C. 

72.  Elimination  by  comparison  is  a  third  method,  which  con- 
sists in  expressing  the  same  variable  in  terms  of  the  other  in  each 
equation  and  equating  these  two  expressions  to  each  other. 

As  an  example  of  elimination  by  comparison,  solve 

i3y  +  x  =  U/  (1) 

\2y-5x  =  -19.  (2) 

From(l),                               x  =  14:-dy.  (3) 

From  (2),  x  =  ll±ll.  (4) 

5 

From  (3)  and  (4),      U  -  3  y  =  ^^-±ll .  (5) 

o 
Solving  (5),  y  =  3. 

Substituting  in  (1),  x  =  5. 

Check  by  substituting  x  =  5,  y  =  3  in  both  (1)  and  (2). 

In  applying  any  method  of  elimination  it  is  desirable  first 
to  reduce  each  equation  to  the  standard  form ;  ax-{-by=c. 
See  §  123,  E.  C. 

EXERCISES 

Solve  the  following  pairs  of  equations  by  one  of  the  pro- 
cesses of  elimination. 


.-44, 

J- 


J     i3x-{-2y  =  118,  2     i5x-Si  =  7y 

'    \x-{-5y=zl91,  '    [2x  =  y-\-^. 

3    I6x-Sy  =  7,  {Sx  +  7y-34.1  =  7iy  +  ^S^^x, 

\2x-2y  =  3,  '    \2lx-^iy  =  l. 


32/4-40=  2a;  4- 14, 
347  =  5  a; -420. 


{5x-lly-2=4.x,  I3y 

'    \5x-2y  =  63.  '    hy 

|52/-3a;  +  8  =  4?/  +  2a;-f-7,  f62/-5a;=:5a;4-14, 

l4a;-*22/=32/  +  2.  *    \3y-2x-6  =  5 -\-a 


318  INTEGRAL  LINEAR  EQUATIONS 

o     |(^+5)(2/+7)=(a:+l)(2/-9)+112,     ^^     \7S-7y  =  5x, 
l2a;  +  10  =  32/+l.  [2y-3x  =  12. 

^^     \ax  =  by,  13     \x+y  =  a,  ^^ 

\x  +  y  =  c.  [x—'y  =  h. 


12.    ia     b  14. 


\ax  +  by  =  Cj  ^q 


[fx  +  gy  =  h. 


3 

_5_ 

=  6, 

X 

y 

2 

+5= 

:2. 

X 

y 

a 

,  b 

+  -  = 

:C, 

\x 

^ 

f 

+  ^  = 

:^. 

.»     2/ 


SOLUTION  BY  FORMULA 
73.  We  now  proceed  to  a  more  general  study  of  a  pair  of 
linear  equations  in  two  variables. 

^     ,     ^,  j2a:  +  32/  =  4,  (1) 

Ex.    1.      Solve  \   ^  n  rr  /0\ 

[5x  +  ey  =  7.  (2) 

Multiplying  (1)  by  5  and  (2)  by  2, 

5-2a:  +  5-  32/  =  5  .4,  (3) 

2-5x  +  2'Qy  =  2  -7.  (4) 

Subtracting  (3)  from  (4),  (2  •  6  -  5  •  3)y  =  2  .  7  -  5  •  4.  (5) 

Solving  for  y,  y  =  l[lzl',l  =  ff  =  ^^  ^ 

In  like  manner,  solving  for  x  by  eliminating  y, 

4.6-7.3        3  .  ,7x 

we  have  ^  =_____==_=:- 1.  (7) 

Ex.   2.    In  this  manner,  solving, 

(7x-^9y  =  71, 
l2a;  +  3  2/  =  48, 

-    ,  7I.3I48.9  ^    ,  „     7  .  48  -  2  .  71 

^^^^^  ^=   7.3-2.9    ""^  ^=  7.3-2.9- 

*  Let  -  and    -  be  the  unknowns.  • 


SOLUTION  BY  FORMULA  319 

In  Ex.  2,  the  various  coefficients  are  found  to  occupy  the 
same  relative  positions  in  the  expressions  for  x  and  y  as  the 
corresponding  coefficients  do  in  Ex.  1. 

Show  that  this  is  also  true  in  the  following : 

Ex.3.    |3^  +  ^2/  =  10,        Ex.4.    |5a^-32/  =  8, 
I2x  —  5y  =  7.  I2x  +  7y  =  19. 

A  convenient  rule  for  reading  directly  the  values  of  the  un- 
knowns in  such  a  pair  of  equations  may  be  made  from  the 
solution  of  the  following: 

Ex.5.     Solve  j«i»'  +  6.3/  =  c„ 

[a^  -{-00  =  Cg. 

Eliminating  first  y  and  then  x  as  in  Ex.  1,  we  find: 
^  ^  C]^>2  -  ^2^1  and  y  =  "1^2  ~  "2^1. 


To  remember  these  results,  notice  that  the  coefficients  of  x  and  y  in 
the  given  equations  stand  in  the  form  of  a  square,  thus    ^    ^  and  that 

the  denominator  in  the  expressions  for  both  x  and  y  is  the  cross 
product  a^2  minus  the  cross  product  a^y  The  numerator  in  the 
expression  for  x  is  read  by  replacing  the  a's  in  this  square  by  the  c's, 

c  h 
i.e.,    ^  \  and  then  reading  the  cross  products  as  before.     The  numera- 

tor  for  y  is  read  by  replacing  the  6's  by  the  c's,  i.e.,  ^  \  and  then 
reading  the  cross  products.  -   ^ 

74.   To  indicate  that  the  coefficients  in  a  pair  of  equations  are 


to  be  treated  as  just  aescribed,  we  write 


=  tti^g  —  ot2^i  ^iid 


call 


a^  &2 
a  determinant.   These  are  much  used  in  higher  algebra. 


Since  any  pair  of  linear  equations  in  two  unknowns  may  be 
reduced  to  the  standard  form  as  given  in  Ex.  5,  it  follows  that 
the  values  of  x  and  ?/  there  obtained  constitute  a  formula  for 
the  solution  of  any  pair  of  such  equations. 


320  INTEGRAL  LINEAR  EQUATIONS 

EXERCISES 

Reduce  the  following  pairs  of  equations  to  the  standard  form. 
Solve  by  use  of  the  formula  Exs.  1-7,  and  12-14. 
^     f3aj  +  42/  =  10,  g     {ax-by  =  0, 

'    14  a; +  2/ =  9.  '    \x  —  y=  c. 

j  4  ic  —  5  2/  =  —  26,  imx  +  ny=p, 

[2  X  ~  3  y  =  —  14:.  '    \rx  +  sy  =  t. 

{6y-17  =  -5x,  ia(x-{-y)-h(x-y)=z2ay 

\6x-16=-5y.       '  \a{x-y)-b(x  +  y)  =  2b. 

a(x-S)=-i(y-2)+ix,  a7c  +  l)x-{-(Jc-2)y  =  Sa, 

li(2/_l)=aj-i(a;-2).       ^-    \(k +3)x-h(7c-4:)y  =  7 a. 

{2x-y  =  53,  {2ax  +  2hy  =  4a'  +  h\ 

ll9x+172/  =  0.  {x-2y=2a--h. 

Ua  -\-h)x  -  {a  -h)y  =  4.  ah, 
^^'    l(a-&)a;+(a  +  6)2/  =  2a2-2  62. 

i„     a{a-h)-\{a-3h)  =  h-3, 
^'"    l3(a_6)  +  |(a  +  6)  =  18. 


2. 


3. 


4. 


"^    '''x-\-y)-h\x-y)=a 
1,.    .7(.-5)=3-|-., 


i(»-2/)+i2/-|(»'-l)=-l- 
r  mo;  4-  71?/  =  m^  +  2  m^n  +  w^, 
1  na;  +  my  =  m^  +  2  mn^  +  w^. 

r  (m  4-  n)x  —  (m  —  n)y  =  2  ?m, 
I  (m  +  Z)a;  —  (??i  —  l)y  =  2  mn. 


17. 

18. 


f  (a  —  b)x-{-(a  +  %  =  2a, 


(a  -  b)x  —  {a  +  b)y  =  2  b. 

{h  +  k)x  +  (/i  -  l€)y  =  2(/i2  +  A^, 

(/i  -  A:)a;  +  (/i  +  k)y  =  2 {h^  -  A;-). 

(a_6)aj  +  2/(a24-&')  =  («  +  '^)'  +  «  +  ^-2«^^ 


r(a-6)x  +  2/(a^4-&0  =  («  +  '^r  +  «  +  ^ 


INCONSISTENT  AND   DEPENDENT  EQUATIONS         321 


INCONSISTENT    AND    DEPENDENT   EQUATIONS 

75.  A  pair  of  linear  equations  in  two  variables  may  be  sucli 
that  they  either  have  no  solution  or  have  an  unlimited  number 
of  solutions. 

r   x-2y  =  -2,  (1) 

l3a;-6y  =  -12,  (2) 


Ex.  1.     Solve 


On  graphing  these  equations  they  are  found  to  represent  two  par- 
allel lines.  Since  the  lines  have  no  point  in  common,  it  follows  that 
the  equations  have  no  solution.     See  Fig.  1. 

Attempting  to  solve  them  by  means  of  the  formula,  §  73,  we  find  : 

12 


0 


and 


^(-2)(-6)-C-12)(-2)^-12 
l(_6)-3(-2) 

^  l(-12)-3(-2)  ^  -Q^ 
l(_6)-3(-2)        0  • 


6 


But  by  §  25, and  are  not  numbers.     Hence,  from  this  it 

follows  that  the  given  equations  have  no  solution.     In  this  case  no 
solution  is  possible,  and  the  equations  are  said  to  be  contradictory. 


Fig.  1. 


Fig.  2. 


3fl;  — 6i/  =  — 6, 


(1) 
(2) 


Ex.2.     Solve  ^6x-^y^-^ 

I    x-2y  =  -2, 

On  graphing  these  equations,  they  are  found  to  represent  the  same 
line.  Hence  every  pair  of  numbers  satisfying  one  equation  must 
satisfy  the  other  also.     See  Fig.  2. 


322.         INTEGRAL  LINEAR  EQUATIONS 

Solving  these  equations  by  the  formula,  we  find : 

^^(-6)(-2)-(-2)(-6)^0^^  3(-2)-lC-6)^0 

3(_1>)_1(_6)  0  ^      a(-2)~l(-6)      O' 

But  by  §  24,  -  may  represent  any  number  ivhatever.     Hence  we  may 

select  for  one  of  the  unknowns  any  value  we  please  and  find  from  (1) 
or  (2)  a  corresponding  value  for  the  other,  but  we  may  not  select 
arbitrary  values  for  both  x  and  y. 

In  this  case  the  solution  is  indeterminate  and  the  equations 
are  dependent;  that  is,  one  may  be  derived  from  the  other. 
Thus,  (2)  is  derived  from  (1)  by  dividing  both  members  by  3. 

76.  Two  linear  equations  in  two  variables  which  have  one 
and  only  one  solution  are  called  independent  and  consistent. 

The  cases  in  which  such  pairs  of  equations  are  dependent  or  contra- 
dictory are  those  in  which  the  denominators  of  the  expressions  for  x 
and  y  become  zero.  Hence,  in  order  that  such  a  pair  of  equations 
may  have  a  unique  solution,  the  denominator  a^ftg ""  "2^1  ^^  *^® 
formula,  §  73,  must  not  reduce  to  zero.  This  may  be  used  as  a  test  to 
determine  whether  a  given  pair  of  equations  is  independent  and  con- 
sistent. 

EXERCISES 

In  the  following,  show  both  by  the  formula  and  by  the  graph 
which  pairs  of  equations  are  independent  and  consistent,  which 
dependent,  and  which  contradictory. 

^    I5x-Sy  =  5,  i3x-6y+5  =  2x-5y-j-T, 

'    l5a;-32/  =  9.  '  \5x+Sy-l  =  3x+5y-^3. 

^    ^x-7-{-5y  =  y-x-2,  ^     (2y-\-7x  =  2  +  6x, 

l5x-\-3y  —  4:  =  dx-y+S.       '   \4:X  —  Sy  =  4. -}- 3x- 


5y, 

2x-7.  '     l2a;  +  7  =  42/-9. 


3    j^7x-3y-4.==2x-2,  ^    (5x-3  =  7y  +  S, 

lx-3y  =  6,  (5x  +  2y==6-\-3x  +  5y, 

■    I5a;-152/  =  18.  '    \3x-}-y  =  lS-3x-{-10y. 

^     (3y-Ax-l  =  2x-5y+S,^^    j 3x-\-4:y  =  7 +  5y, 
^2y-5x-^S  =  3x-^y.  '    ix~y=^6-2x. 


EQUATIONS   IN  MORE   THAN   TWO   UNKNOWNS        323 

SYSTEMS   OF   EQUATIONS   IN   MORE   THAN   TWO   UNKNOWNS 

77.  If  a  single  linear  equation  in  three  or  more  variables  is 
given,  there  is  no  limit  to  the  number  of  sets  of  values  whicli 
satisfy  it. 

E.g.  3a:  +  2y  +  4:2  =  24is  satisfied  by  a;  =  1,  ^  =  3,  z  =  3f ;  x  =  2, 
y  =  2,  3  =  3^ ;  a:  =  0,  y  =  0,  z  =  6 ;  etc. 

If  two  linear  equations  in  three  or  more  variables  are  given, 
they  have  in  general  an  unlimited  number  of  solutions. 

E.g.  3a;  +  2y  +  4:z  =  24  and  a:  +  y  +  z  =  6  are  both  satisfied  by 
a;  =  2,  y  =  -  I,  z  =  5 ;  a;  =  3,  y  =  —  1^,  z  =  4J ;  etc. 

But  if  a  system  of  linear  equations  contains  as  many  equa^ 
tions  as  variables,  it  has  in  general  one  and  only  one  set  of 
values  which  satisfy  all  the  equations. 

E.g.  The  system   J3x-3/  +  2z  =  7, 
l2a;  +  3y-z  =  5, 
is  satisfied  by  a:  =  1,  y  =  2,  z  =  3,  and  by  no  other  set  of  values. 

It  may  happen,  however,  as  in  the  case  of  two  variables, 
that  such  a  system  is  not  independent  and  consistent. 

Such  cases  frequently  occur  in  higher  work,  and  a  general  rule  is 
there  found  for  determining  the  nature  of  a  system  of  linear  equa- 
tions without  solving  them  ;  namely,  by  means  of  determinants  (§  73).  In 
this  book  the  only  test  used  is  the  result  of  the  solution  itself  as 
explained  in  the  next  paragraph. 

78.  An  independent  and  consistent  system  of  linear  equa-. 
tions  in  three  variables  may  be  solved  as  follows : 

From  two  of  the  equations,  say  the  1st  and  2d,  eliminate  one  of 
the  variables,  obtaining  one  equation  in  the  remaining  two  variables. 

From  the  1st  and  3d  equations  eliminate  the  same  variable,  ob- 
taining a  second  equation  in  the  remaining  two  variables. 

Solve  as  usual  the  two  equations  thus  found.  Substitute  the 
values  of  these  two  variables  in  one  of  the  given  equations,  and  thus 
find  the  value  of  the  third  variable. 

The  process  of  elimination  by  addition  or  subtraction  is  usually 
most  convenient.    See  §  125,  E.  C. 


324 


INTEGRAL  LINEAR   EQUATIONS 


EXERCISES 


Solve  each  of  the  following  systems  and  check  by  substitut- 
ing in  each  equation : 


(2x-j-5y-7z  =  9, 
1.    1 5  a;  —  2/  +  3  2!  =  16, 
[7  x+6y+z  =34:. 


rSz-Sy  +  x=-2, 
3x-5y-6z  =-46, 
[y  -\- 5  x—4.z=  —IS. 


'a  +  6  +  c  =  9, 

8a  +  46  +  2c  =  36, 

27a  +  96+3c  =  93. 


*  Use  -,  T,  and  -  as  the  unknowns. 
a  0  c 


6/ 


'3 

2 

~? 

a 

2 

,  5 

4 

+  7- 

a 

b 

c 

I 

3  , 

6 

-tH- 

la 

b 

c 

=  17, 


rl8  ?  -  7  m  -  5  w  =  161, 
3.    U^m  —  ^l -{-71  =  1S, 
[3J  n  +  2  m  +  f  Z  =  33. 


ra;  +  2?/-32J=—  3, 
[5£c  —  4?/  —  72;  =  — 5. 


3  +  6^9"      ^' 


^6      9^12  ^' 


a  .   b        c  _ 
9     12     15"" 


fa; +  2/ =  16, 

I.    |2  +  a;  =  22, 

[2/  +  2  =  28. 


p  aj  4-  3  2/  _  7  2?  =  19, 
8.    is  a; +  8?/ +  11^=24, 
[7  a;  +  11 2/  +  4  2;  =  43. 

Show  that  this  system  is  not 
independent. 


{a  4-  6  +  c  =  5, 
3  a  _  5  6  +  7  c  =  79, 
9  a  -  11  6  =  91. 


raj  +  22/  =  26, 
10.    J3aj  + 42  =  56, 
[5  j^  +  6  2!  =  65. 


(l-\-m-}-n  =  29J, 

12.    h  +  m-n  =  18J, 

i?  —  m  +  n  =  13J. 


13. 


14. 


EQUATIONS  IN  MORE  THAN  TWO   UNKNOWNS        325 


r?  +  m  +  n  =  a, 
Z  +  m  —  n  =  5, 
[l  —  m  -^  n  =c. 

cy  +  dz  =  q, 
ex  ■\-fz  =  r. 


15. 


a     0 


4, 


a     c       ' 
b      c 


16. 


17. 


w-f    v-{-    x+   y  =  10j 

Su-\-4:V  —  5x  +  6y  =  20j 
.4:tc-\-5v-{-(ix  —  7y  =  4:. 


-  +  -  =  a, 
X     y 

X     z        ' 


18.   Make  a  rule  for  solving  a  system  of  four  or  more  linear 
equations  in  as  many  variables  as  equations. 


PROBLEMS  INVOLVING  TWO  OR  MORE  UNKNOWNS 

1.  A  man'  invests  a  certain  amount  of  money  at  4%  inter- 
est and  another  amount  at  5%,  thereby  obtaining  an  annual 
income  of  $3100.  If  the  first  amount  had  been  invested  at 
5%  and  the  second  at  4%,  the  income  would  have  been 
^3200.     Find  each  investment. 

2.  The  relation  between  the  readings  of  the  Centigrade 
and  the  Fahrenheit  thermometers  is  given  by  the  equation 
F  =  32  +  I C.  The  Fahrenheit  reading  at  the  melting  tem- 
perature of  osmium  is  2432  degrees  higher  than  the  Centigrade. 
Find  the  melting  temperature  in  each  scale. 

In  the  Reaumur  thermometer  the  freezing  and  boiling  points  are 
marked  0°  and  80°  respectively.  Hence  if  C  is  the  Centigrade  reading 
and  R  the  RIaumur  reading,  then  R  =  |  C. 

3.  What  is  the  temperature  Fahrenheit  (a)  if  the  Fahren- 
heit reading  equals  i  of  the  sum  of  the  other  two,  (b)  if  the 
Centigrade  reading  equals  -J-  of  the  Fahrenheit  minus  the  Reau- 
mur, (c)  if  the  Eeaumur  is  equal  to  the  sum  of  the  Fahrenheit 
and  Centigrade  ? 


326  INTEGRAL  LINEAR  EQUATIONS 

4.  Going  with  a  current  a  steamer  makes  19  miles  per 
hour,  while  going  against  a  current  J  as  strong  the  boat  makes 
5  miles  per  hour.     Find  the  speed  of  each  current  and  the  boat. 

5.  There  is  anumber  consisting  of  3  digits  whose  sum  is  11. 
If  the  digits  are  written  in  reverse  order,  the  resulting  number 
is  594  less  than  the  original  number.  Three  times  the  tens'  digit 
is  one  more  than  the  sum  of  the  hundreds'  and  the  units'  digit. 

6.  A  certain  kind  of  wine  contains  20  %  alcohol  and  another* 
kind  contains  28  %.  How  many  gallons  of  each  must  be  used 
to  form  50  gallons  of  a  mixture  containing  21.6  %  alcohol  ? 

7.  The  area  of  a  certain  trapezoid  of  altitude  8  is  68.  If  4 
is  added  to  the  lower  base  and  the  upper  base  is  doubled,  the 
area  is  108.     Find  both  bases. 

A  trapezoid  is  a  four-sided  figure  whose  upper  base,  h^,  and  lower 
base,  1)2,  are  parallel,  but  the  other  two  sides  are  not.  If  A  is  the  per- 
pendicular distance  between  the  bases,  then  the  area  is  a  =  -  (bi  +  b2). 

8.  The  Centigrade  reading  at  the  boiling  point  of  alcohol 
is  96°  lower  than  the  Fahrenheit  reading.  Find  both  the 
Centigrade  and  the  Fahrenheit  reading  at  this  temperature. 

Use  C  and  F  as  the  unknowns.  Then  one  of  the  equations  is  the 
formula  connecting  Fahrenheit  and  Centigrade  readings  obtained  on 
page  313  and  the  other  is  C  +.96  =  F. 

9.  The  Fahrenheit  reading  at  the  temperature  of  liquid 
air  is  128  degrees  lower  than  the  Centigrade  reading.  Find 
both  the  Centigrade  and  the  Fahrenheit  reading  at  this 
temperature. 

10.  The  Centigrade  reading  at  the  melting  poiSt  of  silver 
is  796°  lower  than  the  Fahrenheit  reading.  Find  both  Centi- 
grade and  Fahrenheit  readings  at  this  temperature. 

11.  The  Fahrenheit  reading  at  the  melting  point  of  gold  is 
992°  higher  than  the  Centigrade  reading.  Find  both  Centi- 
grade and  Fahrenheit  readings  at  this  temperature. 


PROBLEMS   IN   TWO   OR   MORE   UNKNOWNS  327 

12.  The  upper  base  of  a  trapezoid  is  6  and  its  area  is  168. 
if  I  the  lower  base  is  added  to  the  upper,  the  area  is  210. 
Find  the  altitude  and  the  lower  base. 

13.  A  and  B  can  do  a  piece  of  work  in  18  days,  B  and  C  in 
24  days,  and  C  and  A  in  36  days.  How  long  will  it  require 
each  man,  working  alone,  to  do  it,  and  how  long  will  it  require 
all  working  together  ? 

14.  A  and  B  can  do  a  piece  of  work  in  m  days,  B  and  C  in  n 
days,  and  C  and  A  in  p  days.  How  long  will  it  require  each  to 
do  it  working  alone? 

15 .  A  beam  resting  on  a  fulcrum  balances  when  it  carries 
weights  of  100  and  130  pounds  at  its  respective  ends.  The 
beam  will  also  balance  if  it  carries  weights  of  80  and  110 
pounds  respectively  2  feet  from  the  ends.  Find  the  distance 
from  the  fulcrum  to  the  ends  of  the  beam. 

16.  A  beam  carries  three  weights.  A,  B,  and  C.  A  balance 
is  obtained  when  A  is  12  feet  from  the  fulcrum,  B  8  feet  from 
the  fulcrum  (on  the  same  side  as  A),  and  C  20  feet  from  the 
fulcrum  (on  the  side  opposite  A).  It  also  balances  when  the 
distance  of  ^  is  8  f/et,  B  10  feet,  and  O  18  feet.  Find 
the  weights  B  and  G  it  A  is  50  lbs. 

17.  At  0°  Centigrade  sound  travels  1115  feet  per  second 
with  the  wind  on  a  certain  day,  and  1065  feet  per  second 
against  the  wind.  Find  the  velocity  of  sound  in  calm  weather, 
and  the  velocity  of  the  wind  on  this  occasion. 

18 .  If  the  velocities  of  sound  in  air,  brass,  and  iron  at  0° 
Centigrade  are  x*,  y,  z  meters  per  second  respectively,  then 
3a; -1-2?/- 2  =  2505,  5x-2y  +  z  =  Wl,  and  x  +  y  +  z  =  8777. 
Find  the  velocity  in  each. 

19.  If  ic,  y,  z  are  the  Centigrade  readings  at  the  temperatures 
which  liquefy  hydrogen,  nitrogen,  and  oxygen  respectively, 
then  Sx  —  Sy  +  2z  =  U0,  —  Sx-{- 2 y -\- 4:Z  =  903,  and  x-^Ay 
—  6z  =  60.  Find  each  temperature  in  both  Centigrade  and 
Fahrenheit  readings. 


328  INTEGRAL  LINEAR  EQUATIONS 

20.  If  X,  y,  z  are  the  Centigrade  readings  at  the  freezing 
temperatures  of  hydrogen,  nitrogen,  and  oxygen  respectively, 
then  we  have  x-^  y  —  3z  =  199,  2x  —  5y  -\-  z  =  328,  and 
^4:X-\-2y-{-2z  =  156.    Find  each  temperature. 

21.  li  X,  y,  z  are  respectively  the  melting  point  of  carbon, 
the  temperature  of  the  hydrogen  flame  in  air,  and  the  tempera- 
ture of  this  flame  in  pure  oxygen,  then  10  .t  + 21/  + 2!  =  41,892, 
15a)4-2/4-2;s  =  60,212,  and  7 a; -f  y -f  2  =  29,368.     Eind  each. 

22.  Two  boys  carry  a  120-pound  weight  by  means  of  a  pole, 
at  a  certain  point  of  which  the  weight  is  hung.  One  boy 
holds  the  pole  5  ft.  from  the  weight  and  the  other  3  ft.  from 
it.     What  proportion  of  the  weight  does  each  boy  lift  ? 

Solution.  Let  x  and  y  be  the  required  amounts,  then  5  a:  is  the 
leverage  of  the  first  boy  and  3  y  that  of  the  second,  and  these  must  be 
equal  as  in  the  case  of  the  teeter,  p.  98,  E.  C.     Hence  we  have 

6  a?  =  3  y,  and  a;  +  y  =  120. 

Solving,  we  find  x  =  45,  y  =  75. 

23.  If,  in  problem  22,  the  boys  lift  Fi  and  P2  pounds  respec- 
tively at  distances  di  and  c?2,  and  w  is  the  weight  lifted,  then 

P,d,=.P^  (1) 

P,^F,^w.  (2) 

Solve  (1)  and  (2),  (a)  when  Pj  and  P^  are  unknown,  (6)  when 
Pi  and  w  are  unknown,  (c)  when  Pj  and  dg  are  unknown. 

24.  A  weight  of  540  pounds  is  carried  on  a  pole  by  two  men  at 
distances  of  4  and  5  feet  respectively.    How  much  does  each  lift  ? 

25.  A  weight  of  470  pounds  is  carried  by  two  men,  one  at  a 
distance  of  3  feet  and  the  other  lifting  200  pounds.  At  what 
distance  is  the  latter  ? 

26.  Two  men  are  carrying  a  weight  on  a  pole  at  distances  of 
4  and  6  feet  respectively.  The  former  lifts  240  pounds.  How 
many  pounds  are  they  carrying  ? 


CHAPTER  V 
FACTORING 

79.  A  rational  integral  expression  is  said  to  be  completely 
factored  when  it  cannot  be  further  resolved  into  factors  which 
are  rational  and  integral.     Such  factors  are  called  prime  factors. 

The  simpler  forms  of  factoring  are  given  in  the  following 
outline. 

A  monomial  factor  of  any  expression  is  evident  at  sight, 
and  its  removal  should  be  the  first  step  in  every  case. 

E.g.  4ax^+2  a^x  =  2  ax(2  x -{■  a). 

FACTORS  OF  BINOMIALS 

80.  The  difference  of  two  squares. 

E.g.  4a;2-924  =  (2x -{- Sz^)(2  x  -  Sz^). 

81.  TJie  difference  of  two  cubes. 

E.g.    8x8  _  27^3  =  (O^  -  3y)[(2x)2+  (2x)(Sy)  +  (S  y)^ 
=  (2x-Sy){4x^  +  6xy  +  9y^). 

82.  The  sum  of  two  cubes. 

E.g.      27  a;8  +  64/  =  (3  a:  +  4 y) [(3  xy-  (3  x) (4 y)  +  (4 y)^ 
=  (3x  +  43^)(9x2  -  I2xy  +  IQy^). 

FACTORS  OF  TRINOMIALS 

83.  Trinomial  squares. 

E.g.  a2  +  2  a6  +  &2  =  (a  +  by  =  (a  +  6)(a  +  b), 

and  a'^-2ab  -\-b^=  (a-  by  =  (a  -  b){a  -  6). 

329 


330  FACTORING 

84.  Trinomials  of  the  form  ic^  -\-px  +  q. 

E.g.  a:2  +  3a:-  10=  (a:  +  5)(x-2). 

A  trinomial  of  this  form  has  two  binomial  factors,  x  -\-  a  and  x  +  6, 
if  two  numbers  a  and  h  can  be  found  whose  product  is  q  and  whose 
algebraic  sum  is  p. 

85.  Trinomials  of  the  form  my?  ■\-nx-\-r. 

E.g.     '  6x2  +7a:-20=  (3a:-4)(2x  +  5), 

A  trinomial  of  this  form  has  two  binomial  factors  of  the  type 
ax  ■\-h  and  ex  +  rf,  if  four  numbers,  a,  b,  c,  d,  can  be- found  such  that 
ac  =  m,  bd  =  r,  and  ad  +  be  =  n.     See  §  155,  E.  C. 

86.  Trinomials  which  reduce  to  the  difference  of  two  squares. 

E.g.    x^  +  x^y^  -\-  y^  =  x^  +  2  x^  -h  y*  -  xhf-  =  {f-  +  y'^Y  -  x^y^ 
=  (a:2  +  ?/2  -  xy)  (x^  +  2/2  +  xy). 

In  this  case  x^y"^  is  both  added  to  and  subtracted  from  the  expres- 
sion, whereby  it  becomes  the  difference  of  two  squares.  Evidently 
the  term  added  and  subtracted  must  itself  be  a  square,  and  hence  the 
degree  of  the  trinomial  must  be  4  or  a  multiple  of  4,  since  the  degree 
of  the  middle  term  is  Aa(/'that  of  the  trinomial. 

Ex.        4  a8  -  16  a4&4  +  9  &«  =  4  a^  -  12  a'^h^  +  9  68  -  4  a^J* 
=  (2a4-3M)2-4a464 
=  (2  a*  -  3  6^  +  2  a2^>2)(2a4  _  3  ^4  _  ^a'^b"^). 

EXERCISES  ON  BINOMIALS  AND  TRINOMIALS 

Factor  the  following : 

1.  a^  +  h\                   5.  7aic2-56aV.  9.  ^^^--i^rsK 

2.  a^-h\                    6.  a^-ab\  10.  Sr^-27r. 

3.  (a  +  by-c\           7.  121x^-^xy\  11.  {a-{-hf-c\ 

4.  (a  +  6)3  +  c^  8.   ict'^  +  rb^'-  ^'^'    c' -(«-&)'- 

13.  nc^+1  cd-Qd\  15.   4ar^-12a^?/  +  9/. 

14.  x^-Zx^y'^  +  y\  16.   a^  +  H  ^^  +  30  2^. 


FACTORS   OF  TRINOMIALS  331 

17.  6x^-5xy-6y^.  24.  a^  +  lOa-SO. 

18.  3aV2/*-69aV+336a2.  25.  Say-^Sayz  +  72a^yz\ 

19.  20  a'b^-{- 23  abx- 21  x".  26.  4m8-60mV  +  8lM^ 

20.  a^  +  2a262  4.9  6^  27.  35  a-* -  6  a^ft^' -  9  fo^*. 

21.  ASa^x'y~75af.  28.  (aH- 6)2-(c-d)2. 

22.  16aV?/  +  54a2/^  29.  72  aV-19aV- 40?/^ 

23.  ajy  +  2a^y;2  +  2'.  30.  4(a-3)«-376-(a-3)3+96*. 

31.  6(x  +  yy+5ix^-y^-Hx-yy. 

32.  9(x  - af-  2'i(x  -a)(x-^  a)  +  16(a;  +  af. 

33.  12{c  +  dy-  7(c-{-d)(c-d)-12(c-dy. 

34.  (a2  +  5a-3)2-25(a2  +  5a-3)+150. 

FACTORS   OF  POLYNOMIALS  OP  POUR  TERMS 

A  polynomial  of  four  terms  may  be  readily  factored  in  case 
it  is  in  any  one  of  the  forms  given  in  the  next  three  paragraphs. 

87.  It  may  be  the  cube  of  a  binomial. 
Ex.1.     a^-Sa'b  +  3ab'-b^={a-by. 

Ex.  2.     Sx" +  36  x'y-{-54:xy^ -{-27  f 

=  (2xy  +  S(2xy(3y)  +  3(2:r)(3^)2  +  (3y)8 
=  (2x  +  dyy.    See  Ex.  34,  (d),  p.  23. 

88.  It  may  be  resolvable  into  the  difference  of  two  squares. 
•     In  this  case  three  of  the  terms  must  form  a  trinomial  square. 

Ex.  1.     a^  -  c"  +  2  ah  ^W  =  {a-  +  2  ab  -[-b^  -  c^ 
=  (a  +  by-c^=(a  +  b-h  c)(a  +  b-  c). 

Ex.2.     4.x''-\-z^-4.x*-l  =  z^-4x'  +  4.x^-l 
=  26  _  (4  x*  -  4  a:2  +  1)  =  26  _  (0  3^2  _  1)2 

=  (28  +  2a:2  -  i)(-2:3  _  2 x^  +  1). 


332  FACTORING 

89.  A  binomial  factor  may  be  shown  by  grouping  the  terms. 

In  this  case  the  terms  are  grouped  by  twos  as  in  the  following 
examples. 

Ex.  1.   ax-{-ay-\-  ba?  +  bxy  =  (ax  -\-  ay)  +  (bx^  +  bxy) 
=  a(x  +  y)  +  bx{x  -\-  y)  =  (a  +  hx)  {x  +  y). 

Ex.2.   ax  +  bx  +  a''-b''=(ax-\-bx)  +  {a^-b') 

=  x{a  +  J)  +  (a  -  h){a  +  &),=  {x  +  a  -  b)(a  +  b). 

EXERCISES 

Factor  the  following  polynomials: 

1.  x^  +  Sx'y-\-3xy'^  +  f.  8.  a^b^ -  a^bc'n - abn  +  awV. 

2.  8a3-36a26  +  54a62_27&3.      9.  2/  +  46?/  +  3c2/  +  6&c. 

3.  4a4-4a-6=^  +  6^-16a;2.  10.  6c?/2  +  c^^^  +  ftt^^/ +  c^c;?. 

4.  2ad  +  3bc-{-2ac  +  Sbd.  11.  5 a^c  + 12 cd- 6 ac?- 10 ac^. 

5.  2To(^-54.x'y  +  S6xy''-Sf.  12.  a^ - 6 V  +  aca^ - ftcx^. 

6.  36a^-24a«4-24a-16.  13.  b^c^-c'f-bY-hy'. 

7.  mna^  —  mrcc  —  7'n^ic  +  ?'-w.  14.  a^  —  2  a^*6*  —  2  a*6^*  +  6^. 

15.  m'*+^  +  m"7l*  +  m*/^^^-7^"+^ 

1 6.  by  -b{c-d)f-  +  d{by-c)  +  d?. 

FACTORS  OF  POLYNOMIAL  SQUARES 

90.  In  §  158,  E.  C,  squares  of  trinomials  were  factored  by 
inspection.  In  the  examples  there  considered  there  were  no 
similar  terms  combined  in  the  polynomial  squares.  When  this 
is  the  case,  it  is  easy  to  recognize  at  sight  any  polynomial 
square  and  hence  to  get  its  square  root. 

E.g.  a^  +  b^+c'^-{-d-2  +  2ab  +  2ac  +2  ad  +  2bc  +  2bd  +  2cd  is  the 
square  ota  +  b  +  c  +  d. 

It  is  also  possible  in  some  cases  to  recognize  a  polynomial 
square  after  some  similar  terms  have  been  combined. 

E.g.  Examine  x^  -\-  2  x^  +  '^  x^  +  2  x  -\- 1.  We  see  that  x^  and  1  are 
squares.     If  this  polynomial  is  a  square,  there  must  be  another  squared 


FACTORS  FOUND   BY  GROUPING  333 

term  which  is  combined  with  some  other  term.  3  x^  is  the  only  such 
term  possible,  and  hence  the  other  square  must  be  x^.  Hence,  we 
try  a:^  a;^,  and  1  for  the  squared  terms  and  soon  see  that  x'^  ^x-\-\  is 
the  required  square  root. 

EXERCISES 

Factor  the  following : 
1.   4a;*  +  9/  +  l-12an/  +  4«-6?/. 

11^222 

2-    3  +  -2+1+-+-  +  -- 
a^     2/2  xy     X      y 

a^      c2  2ac     4a     4c 

4.  a;^_2ar'  +  3ic2-2a;  +  l. 

5.  x^-2a?-x'-{-2x  +  l. 

6.  9a«  +  12a*-6a»4-4a2-4a4-l. 

7.  a;«-8ar'  +  22a;*-24iB3  +  9a;^. 

8.  a;«  +  8a;^-2i»2_44-i. 

9.  a4_2a3  +  a2  +  2-?  +  -i. 

a     a* 

10.  4a;24.94._i_4-6a;4-2  +  5. 

4  a;-  X 

11.  a2+4Z>2+9c-+d2_4«2,_|.6^c-2ad-126c+46c?-6cd 

FACTORS  FOUND  BY  GROUPING 

91.  The  discovery  of  factors  by  the  proper  grouping  of  terms 
as  in  §  89  is  of  wide  application.  Polynomials  of  five,  six,  or 
more  terms  may  frequently  be  thus  resolved  into  factors. 

Ex.  1.   a2_^2a&  +  62  +  5a  +  56=(a  +  6)-'+(5a  +  56) 
=  (a  +  b)  (a  +  ?;)  +  5(a  +  ft)  =  (a  +  6  +  5)  (a  +  h). 

Ex.  2.   x^-7x-^6-ax-\-6a  =  x'^-7x-i-6-(ax-6a) 
=  (x  -1)  (X  -  G)  -  a(x  -  Q)  =  (x  -  Q)  (x  -1  -  a). 

Ex.3.    a'-2ab-[-b'-x'-2xy-f 

=  (a^-2ah  +  b^)-(x^  +  2xy  +  y^) 

=  (a  -  &)2  -  (a;  _  2/)2  =  (a  -  6  +  a;  -  y)  (a  -  6  -  a:  +  3^). 


834  FACTORING 

Ex.4.   ax^+ax-6a  +  x^-{-7x-\-12 

=  a(pc-^  +  :c  _  6)  +  (x2  +  7  X  +  12) 
*        =  a(x  +  S)(x  -  2)  +  (a;  +  d)(x  +  4) 

=  (x  +  3)[a(x  -  2)+  X  -\-  4:']  =  (x  +  3)(ax  -2a  +  X  +  4:). 

In  some  cases  the  grouping  is  effective  only  after  a  term  has 
been  separated  into  two  parts. 

Ex.5.   2a^  +  3a''  +  3a  +  l  =  a^  +  {a^  +  Sa^+3a-\-l) 

=  a^+(a  +  iy  =  (a  +  a  +  l)[a2  -a(a  +  l)  +  (a  +  1)2] 
=  (2a  +  l)(a2  +  a  +  l). 

As  soon  as  the  term  2  a^  is  separated  into  two  terms  the  expression  is 
shown  to  be  the  sum  of  two  cubes. 

Again,  the  grouping  may  be  effective  after  a  term  has  been 
both  added  and  subtracted : 

Ex.  6.  a*  +  b'  =  (a^  +  2  a^ft^  ^b*)-2  a-b^ 
=  (a^  +  b^y-(ahV2y 
=  (a2  +  b'^  +  ab  V2)(a2  +  b^  -  aby/2). 

In  this  case  the  factors  are  irrational  as  to  one  coefficient.     Such 
factors  are  often  useful  in  higher  mathematical  work. 

EXERCISES 

Factor  the  following : 

1.  a^  —  2  xy -{- y^  —  ax -{-  ay.  3.  a^  —  W  —  a^  —  ab  —  b^. 

2.  a^-ab-^W-^a^-\-b\  4.  a^-2ab-^W-x^-]-2xy-y\ 

5.  a^-\-2a^b-ah^+a^b''-2ab(?-b^c'. 

6.  a;^-2/'  +  aa;2  +  a2/2-a^-2/2.    7.  a"" +  a%'' -{-b^  ^a^ -^bK 
In  7  group  the  first  three  and  the  last  two  terms. 

8.  a^  —  1  4-  3  a;  —  3  ic^  +  a^.     Group  the  last  four  terms. 

9.  a^_|_a^_|_3a;-f  2/3_2^2_|.32^^ 

Group  in  pairs,  the  1st  and  4th,  2d  and  5th,  3d  and  6th  terms. 

10.  x*-^a^y  —  xy^  —  y^-{-a^  —  'if. 

11.  a^  +  4:a^b  +  6a'b^  +  4.ab^-{-b'~x\ 


THE   FACTOR   THEOREM  335 

12.  x*  +  4:ay^z  —  4:f-\-4:yw-\-4.z^  —  iu^. 

13.  2a^-12b^-^3bd  —  5ab-9()c-6ac-\-2ad. 
Group  the  terms :   2  a^  —  5  ab  —  12  bK 

14.  d'-\-ab-4:ac-2b^  +  4:bc-{-3ad-Sbd. 

15.  a^  +  2  -  3  a.     Group  thus :  (a^  -  1)  +  (3  -  3  a). 

16.  4a^  +  a  — 8aic  — a;  +  4ar^. 

17.  3a2_8a6  +  462  +  2ac-46c. 

18.  x*^  4- 1/.     Add  and  subtract  2  .t^. 

19.  a«  +  2a»63^6«-2a*6-2a6*. 

20.  a^-3  a^  +  4.     Group  thus :  (a^  -  2  a^)  +  (4  -  a^), 

21.  a-c  —  ac^  —  a-6  4- aft-^  —  6x' 4- 6c^. 

22.  a-6  —  a^c  +  6-c  —  ab^  +  ac^  —  6c^. 

23.  3ar'-a^-4a;  +  2.     Add  and  subtract  -2x^. 

24.  2a^-lla^  +  lSx-9.     Add  and  subtract  9 x^. 

FACTORS   FOUND   BY   THE  FACTOR  THEOREM 

92.  It  is  possible  to  determine  in  advance  whether  a  poly 
nomial  in  x  is  divisible  by  a  binomial  of  the  form  x  —  a. 

E.g.   In  dividing  a;*  —  4  a;*  +  7  a;^  —  7  a:  +  2  by  a:  —  2,  the  quotient  is 
found  to  be  a:8  -  2  x2  +  3  a:  -  1. 

Since  Quotient  x  Divisor  =  Dividend,  we  have 

(x  -  2)(a,-8  -  2  a:2  +  3a;  -  l)  =  a:*  -  4a;8  +  7a;2  _  7a:  +  2. 

As  this  is  an  identity,  it  holds  for  all  values  of  x.     For  a:  =  2  the  fac- 
tor (a;  —  2)  is  zero,  and  hence  the  left  member  is  zero,  §  22. 

Hence  for  ar  =  2  the  right  member  must  also  be  zero.     This  is  in- 
deed the  case,  viz. : 

2*  _  4  .  28  +  7  .  22  -  7-  2  +  2  =  16  -  32  +  28  -  14  +  2  =  0. 

Hence  if  ar  -  2   is  a  factor  of  a;*  -  4  x^  +  7  a;^  -  7  a;  +  2,  the  latter 
must  reduce  to  zero  for  a:  =  2. 


336  FACTORING 

93.  In  general  let  D  represent  any  polynomial  in  x.  Sup- 
pose D  has  been  divided  by  a;  —  a  until  the  remainder  no  longer 
contains  x.  Then,  calling  the  quotient  Q  and  the  remainder  R, 
we  have  the  identity 

D=q{x-a)-\-R,  (1) 

which  holds  for  all  values  of  x. 

The  substitution  of  a  for  x  in  (1)  does  not  affect  R,  reduces 
Q(x  —  a)  to  zero,  and  may  or  may  not  reduce  D  to  zero. 

(1)  li  X  =  a  reduces  D  to  zero,  then  0  =  0  +  R.  Hence  R  is  zero, 
and  the  division  is  exact.     That  is,  a;  —  a  is  a  factor  of  D. 

(2)  li  X  =  a  does  not  reduce  D  to  zero,  then  R  is  not  zero,  and  the 
division  is  not  exact.     That  is,  a;  —  a  is  not  a  factor  of  D. 

Hence:  If  a  polynomial  in  x  reduces  to  zero  when  a  particular 
number  a  is  substituted  for  x,  then  x  —  a  is  a  factor  of  the  poly- 
nomial, and  if  the  substitution  of  a  for  x  does  not  reduce  the  poly- 
nomial to  zero,  then  x  —  a  is  not  a  factor. 

This  principle  is  often  called  the  factor  theorem. 

In  applying  the  factor  theorem  the  trial  divisor  must 
always  be  written  in  the  form  x  —  a. 

Ex.  1.   Factor  0^  +  60^ '\-3a^  +  x  +  S. 

If  there  is  a  factor  of  the  form  x  —  a,  then  the  only  possible  values 
of  a  are  the  various  divisors  of  3,  namely  +  1,  —  1,  +  3,  —  3. 

To  test  the  factor  a;  +  1,  we  write  it  in  the  form  a:  —  (—  1)  where 
a  =  —  1.     Substituting  —  1  for  a:  in  the  polynomial,  we  have 

l_6  +  3-l  +  3  =  0. 

Hence  ar  +  1  is  a  factor. 

On  substituting  +  1,  +  3,  —  3  for  x  successively,  no  one  reduces  the 
polynomial  to  zero.     Hence  a;  —  1,  a:  —  3,  a:  +  3  are  not  factors. 

Ex,  2.    Factor  3x^-x^-4.x-{-2. 

If  a:  —  a  is  a  factor,  then  a  must  be  a  factor  of  +  2.  "We  therefore 
substitute,  +2,  —2,  +1,  —1  and  find  the  expression  becomes  zero 
when  +  1  is  substituted  for  x.  Hence  ar  —  1  is  a  factor.  The  other 
factor  is  found  by  division  to  be  3  a;^  +  2  x  —  2,  which  is  prime. 

Hence  3^8_;^2_4^ +  2  =  (a:- l)(3a;2  +  2ar -2). 


THE  FACTOR  THEOREM  837 

EXERCISES 

Factor  by  means  of  the  factor  theorem : 

1.  3a^-2a^  +  5a;-6.  6.   m^  +  Sm'^  +  Tm  +  S. 

2.  2a^H-3a^-3a;-4.  7.   x*  +  Sa^-3a^-T x  +  6. 

3.  2a^-{-x'-12x  +  9.  8.   37^  +  57^-7 r-1. 

4.  a^  +  9a^  +  10ic  +  2.  9.   2s^-\-7 z^  +  4:Z  +  S. 

5.  a3-3a  +  2.  10.   a^-6a^-\-lla-6. 

11.  Show  by  the  factor  theorem  that  x^  —  a*  contains  the 
factor  a;  —  a  if  A;  is  any  integer. 

12.  Show  that  x^  —  a*  contains  the  factor  a?  +  a  if  A;  is  any 
even  integer. 

13.  Show  that  oj*  +  a*  contains  the  factor  a?  +  a  if  A;  is  any 
odd  integer. 

14.  Show  that  a^  +  a*  contains  neither  x-^a  nor  x  —  a  as  a 
factor  if  k  is  an  even  integer. 

MISCELLANEOUS  EXERCISES  ON  FACTORING 

1.  20  aVy  — 45  a^a??/^        4.  16  a;^  -  72  a^  +  81  jr*. 

2.  24  amV- 375  amV.      5.  162a^b-\-252a%^-\-9Sab\ 

3.  432  ar'*s  + 54  ars^        6.  48a^^-12ar'y-12a^y+3y. 

7.  12  a^bx' -{-Sab^x^-\- IS  a'bxy-h  12  ab^xy, 

8.  lSx^y-S9xY  +  lSxf.         16.   a^-f.  17.   a}^-y^\ 

9.  4a;2_93.^_j_g^_92^_^4^_l_g    ^g    a«-f-ay  +  /. 

10.  6ar^-13a;?/  +  6/-3a;H-2y.  19.  o?  +  a-2. 

11.  6a;^  +  15ary  +  92/*.  20.  a8-18aV  +  /. 

12.  16a;^  +  24a:22^2  +  8/.  21.  a^«_6ay  +  /«. 

13.  15a;*  +  24a;2y2^92/*.  22.  o?  +  4:Q?  +  2x-1. 

14.  a^  +  y*.  15.   o}^  +  y^.  23.  3a^  +  2a^-7a;  +  2. 


338  FACTORING 

24.  a^-^aY-\-f.  26.   a^ -{-^ a^ ■\-l%a-\-4:. 

25.  a^  +  a^  +  d  +  l.  27.    2  a;^  +  a^^  +  2  aj^/ +  a;^/^. 

28.  mP  +  m*a  -|-  m^d^  +  9W,W  +  ma*  +  a^. 

29.  (aj-2)3-(2/-^)3. 

30.  a«  +  6^  +  2a6(a*-a262  +  &4). 

31.  xy  +  a;y2  +  ary2;2  +  a;y2;^-l-a72/2;'*  +  2^. 

32.  8a«  +  6a&(2a-36)-2763. 

33.  a  {pa^  +  y^)—ax (o?  —  y^—  y'\x  +  y), 

34.  a3-63  +  362c-36c2  +  c3. 

35.  a*  +  2a3&-2a62c-62c2. 

36.  a*  +  2a36  +  a'&'-a'&'-2a2&2c-6V. 

SOLUTION  OF  EQUATIONS  BY  FACTORING 

94.  Many  equations  of  higher  degree  than  the  first  may  be 
solved  by  factoring.     (See  §§  160-163,  E.  C.) 

Ex.1.   Solve  2ar^-a^-5fl;-2  =  0.  (1) 

Factoring  the  left  member  of  the  equation,  we  have 

(a:-2)(a:+l)(2a:+l)  =  0.  (2) 

A  value  of  x  which  makes  one  factor  zero  makes  the  whole  left 
member  zero  and  so  satisfies  the  equation.  Hence  x  —  2,  a:  =  —  1, 
X  —  —  \  are  roots  of  the  equation. 

To  solve  an  equation  by  this  method  first  reduce  it  to  the  form 
^  =  0,  and  then  factor  the  left  member.  Put  each  factor  equal  to 
zero  and  solve  for  x.  The  results  thus  obtained  are  roots  of  the 
original  equation. 

Ex.2.   Solve  ajs- 12 0^  =  12-35 a;.  (1) 

Transposing  and  factoring,  (a:  —  4)  (a;^  —  8  ar  +  3)  =  0.  (2) 

Hence  the  roots  of  (1)  are  the  roots  of  a;  —  4  =  0  and  a:^  —  8  a;  +  3  =  0. 
From  ar  —  4  =  0,  a:  =  4.  The  quadratic  expiession  a;^  —  8 a:  +  3  can- 
not be  resolved  into  rational  factors.     See  §  155. 


COMMON   FACTORS   AND   MULTIPLES  339 

EXERCI^S 

Solve  each  of  the  following  equations  by  factoring,  obtain- 
ing all  roots  which  can  be  found  by  means  of  rational  factors. 

1.  a^-\-3x'  =  2Sx.  6.  2a^  +  S  x  =  9  x'.-U. 

2.  6x^-\-Sx  +  5  =  19a^.  7.  5  a^  +  a^- 14  aj  +  8  =  0. 

3.  x*-{-12x^-j-3  =  7a^  +  9x,     8.  2  x^-^a^=Sx-3. 

4.  a^  =  -2x^  +  5x  +  6.  9.  a:^  +  2a^  +  12  =  7a^+ 8a;. 

5.  a^-4.x^  =  4:X-\-5.  10.  x* -{- x -{- 6  =  a^ -{- 7 a^. 

COMMON  FACTORS   AND  MULTIPLES 

95.  If  each  of  two  or  more  expressions  is  resolved  into 
prime  factors,  then  their  Highest  Common  Factor  (H.  C.  F.)  is  at 
once  evident  as  in  the  following  example.     See  §  202,  E.  C. 

Given  (1)         x*  -  y*  =  (x^  +  y^)(x  +  y)(x  -  y), 
(2)         x^-y^={x^  +  y^){x^-y^) 

=  (^  +  y)(x2  -xy  +  y^)(x  -  7/Xx^  +  xy  +  y^. 
Then  (x  +  y)(x  -  y)  =  x^  -  y^  is  the  H.  C.  F.  of  (1)  and  (2). 

In  case  only  one  of  the  given  expressions  can  be  factored  by 
inspection,  it  is  usually  possible  to  select  those  of  its  factors^ 
if  any,  which  will  divide  the  other  expressions  and  so  to  deter- 
mine the  H.  C.  F. 

Ex.   Find  the  H.C.F.  of  6a;3 _f_43j2_3a,_2^ 

and  2x^  +  22(^  +  x^-x-l. 

By  grouping  we  find : 

6x8  +  4  j;2  _  3a:  _  2  =  2a:2(3a;  +  2)-(3a;  +  2) 
=  (2a:2-l)(3.r +  2). 

The  other  expression  cannot  readily  be  factored  by  any  of  the 
methods  thus  far  studied.  However,  if  there  is  a  common  factor,  it 
must  be  either  2  a:^  —  1  or  3  a:  +  2.  We  see  at  once  that  it  cannot  be 
3  ar  +  2.  (Why  ?)  By  actual  division  2  z^  -  1  is  found  to  be  a  factor 
oi2x*  +  2x^  +  x^-x-l.    Hence  2^2  -  1  is  the  H.C.F. 


840  FACTORING 

96.  The  Lowest  Common  Multiple  (L.  C.  M.)  of  two  or  more 
expressions  is  readily  found  if  these  are  resolved  into  prime 
factors.     See  §  205,  E.  C. 

Ex.  1.    Given     eabx-6aby  =  2 -S ab(x - 2^),  (1) 

Sa'x-\-Sa'y  =  2^a\x  +  y),  ^   (2) 

36  bXx'  -  y%x  4-  2/)  =  2'm\x  -y){x-\-  y)\  (3) 

The  L.  CM.  is  2^  .  ^'^a%^  {x  -  y){x  +  y)%  since  this  contains  all 
the  factors  of  (1),  all  the  factors  of  (2)  not  fomid  in  (1),  and  all 
the  factors  of  (3)  not  found  in  (1)  and  (2),  with  no  factors  to  spare. 

In  case  only  one  of  the  given  expressions  can  be  factored  by 
inspection,  it  may  be  found  by  actual  division  whether  or  not 
any  of  these  factors  will  divide  the  other  expressions. 

Ex.2.   EindtheL.C.M.  of  6aj3-a;2-|-4a;  +  3,  (1) 

and  6aj^  +  3a;2-10aj-5.  (2) 

(1)  is  not  readily  factored.  Grouping  by  twos,  the  factors  of  (2) 
are  3  a:^  -  5  and  2  a;  +  1.  Now  3  a;2  -  5  is  not  a  factor  of  (1) .  (Why  ?) 
Dividing  (1)  by  2  a;  +  1  the  quotient  is  3  a;^  -  2  a:  +  3. 

Hence       6  a;3  -  a;2  +  4  a:  +  3  =  (2  a:  +  1)(3  a:^  -  2  a;  +  3), 
6a;8  +  3a;2-10a:-5  =  (2  a:  +  1)(3  a;2  -  5). 

Hence  the  L.  CM.  is  (2a:  +  1)(3  a;2  -  2a;  +  3)(3a;2  -  5). 

Ex.  3.  Find  the  L.  C.  M.  of  a^  +  2  a^  -  a  -  2,  (1) 

and  10a^-^o?  +  4.a-{-l,  (2) 

By  means  of  the  factor  theorem,  a  —  1,  a  +  1,  and  a  +  2  are  found 
to  be  factors  of  (1),  but  none  of  the  numbers,  1,  —  1,  —  2,  when  sub- 
stituted for  a  in  (2)  will  reduce  it  to  zero.  Hence  (1)  and  (2)  have 
no  factors  in  common.  The  L.  C  M.  is  therefore  the  product  of  the 
two  expressions ;  viz.  (a  +  l)(a  -  l)(a  +  2)(10a8  -  Sa^  +  4a  +  1). 

97.  The  H.  C.  E.  of  three  expressions  may  be  obtained  by 
finding  the  H.  C.  F.  of  two,  and  then  the  H.  C.  F.  of  this  result 
and  the  third  expression.  Similarly  the  L.  C.  M.  of  three  expres- 
sions may  be  obtained  by  finding  the  L.  C.  M.  of  two  of  them, 
and  then  the  L.  C.  M.  of  this  result  and  the  third  expression. 

This  may  be  extended  to  any  number  of  expressions. 


COMMON  FACTORS   AND   MULTIPLES  341 

EXERCISES 

Find  the  H.  C.  F.  and  also  the  L.C.  M.  in  each  of  the 
following : 

1.  Qi^  +  y^,x^-\-y\  2.  a^  +  xyH-/,  a;'-/. 

3.  x'-^x-Q,   x^-2x-S,   ic2  +  19a;  +  18. 

4.  a;4-6a^  +  l,   a^-\-3^^3x-\-l,   a^  +  Sa^  +  aj-l. 

6.  2x^  +  a^-Sx  +  3,   a^4-2a;-l. 

7.  3r3  +  5r2-7r-l,   3i'^  +  Sr-\-l, 

8.  a3-3a2  +  4,   ax-ab -2 x  +  2b. 

9.  a«^-2a363+6«-2a*6-2a6^   a^ - 2 ab -\- 1^. 

10.  8a3-36a26  +  54a62_276^   4a2-9  6-. 

11.  36a^-9a2-24a-16,  12a^-6a'-Sa. 

12.  22/2  +  4&y4-3ci/H-66c,  / - 3 61/ - 10 6^ 

13.  os^^  —  y^^,   a:^  —  y^,a^  —  y*. 

14.  m'^  +  Sm^  +  Tm,   m^-{-3m^-m-3,  m^-Tm-6. 

98.  An  important  principle  relating  to  common  factors  is 
illustrated  by  the  following  example: 

Given  a;^  +  7  a:  +  10  =  (x  +  5)  (a:  +  2),  (1) 

and  x^-x-6=(x-S)(x  +  2).  (2) 

Add  (1)  and  (2),  2ar«  +  6ar  +  4  =  2(a:  +  l)(a:  +  2).  (3) 

Subtract  (2)  from  (1),  8  a:  +  16  =  8(x  +  2).  (4) 

We  observe  that  x  +  2,  which  is  a  common  factor  of  (1)  and  (2), 
is  also  a  factor  of  their  sum  (3),  and  of  their  difference  (4).  This 
example  is  a  special  case  of  the  following  principle. 


842  FACTORING 

99.  A  common  factor  of  two  expressions  is  also  a  fac- 
tor of  the  swin  or  difference  of  any  multiples  of  those 
expressions. 

For  let  A  and  B  be  any  two  expressions  having  the  common  factor 
/.     Then  if  k  and  I  are  the  remaining  factors  of  A  and  B  respectively, 

A  =fk  and  B  =  fi. 

Also  let  mA  and  nB  be  any  multiples  of  A  and  B. 

Then  mA  =  mfk  and  nB  =  nfl,  from  which  we  have : 

mA  ±nB  =  mfk  ±  nfl  =f(mk  ±  nl). 

Hence /is  a  factor  of  mA  ±  nB. 

100.  Another  important  principle  is  the  following:  If  f  is  a 
factor  of  mA  ±  nB  and  also  of  A,  then  fis  a  factor  of  B,  pro- 
vided n  has  no  factor  in  common  with  A. 

For  let /be  a  factor  of  m^  ±nB  and  also  of  A,  where  m^'and  nB 
:ire  integral  multiples  of  the  expressions  A  and  5. 

'J'hen  /  must  divide  both  mA  and  nB.  We  know  it  divides  mA 
because  it  was  given  as  a  factor  of  ^.  If  it  divides  nB,  it  must  be  a 
factor  of  either  n  or  B.  It  cannot  be  a  factor  of  n,  for  then  A  and  n 
would  have  a  factor  in  common  contrary  to  agreement.  Hence /is  a 
factor  of  B. 

101.  By  successive  applications  of  the  above  principles  it  is 
possible  to  find  the  H.  C.  F.  of  any  two  integral  expressions. 

Ex.  1.   Find  the  H.  C.  F.  of  9  a;^  -  a^  +  2  a;  -  1,  (1) 

and  27a:'-|-8a^-3a;  +  l.  (2) 

Multiplying  (1)  by  3  a;  and  subtracting  from  (2),  we  have 

27  a:5  +  8  a:2  -  3  a;  +  1 
27 a;^-  3ar8  +  Bx^-Sar 

3a;3-H2a:2+l  (3) 

By  §  99,  any  common  factor  of  (1)  and  (2)  is  a  factor  of  (3). 


COMMON  FACTORS  AND   MULTIPLES  343 

Calling  expressions  (1)  and  (2)  B  and  A  respectively  of  principle  2, 
then  (3)  is  ^  —  3  a:  •  i5;  and  since  the  multiplier,  3  a:,  has  no  factor  in 
common  with  (2),  it  follows  from  the  principle  that  any  common 
factor  of  (3)  and  (2)  is  a  factor  of  (1),  and  also  that  any  common 
factor  of  (3)  and  (1)  is  a  factor  of  (2).  Hence  (1)  and  (3)  have  the 
same  common  factors,  that  is,  the  same  H.  C.  F.  as  (1)  and  (2).  There- 
fore we  proceed  to  obtain  the  H.  C.  F.  of 

9a:4_a:2+2x-l,  (1) 

and  3  x8  +  2  a:2  +  1.  (3) 

Multiplying  (3)  by  3  a:  and  subtracting  from  (1)  we  have 

_6a:8-a:2-a:-l.  (4) 

By  argument  similar  to  that  above,  (3)  and  (4)  have  the  same 
H.  C.F.  as  (1)  and  (3)  and  hence  the  same  as  (1)  and  (2).  Multi- 
plying (3)  by  2  and  adding  to  (4)  we  have, 

3  a;2  -  a:  +  1.  (5) 

Then  the  H.  C.  F.  of  (5)  and  (3)  is  the  same  as  that  of  (1)  and  (2). 
Multiplying  (5)  by  x  and  subtracting  from  (3),  we  have 

3  a:2  -  a;  +  1.  (6) 

Then  the  H.  C.  F.  of  (5)  and  (6)  is  the  same  as  that  of  (1)  and  (2). 
But  (5)  and  (6)  are  identical,  that  is,  their  H.  C.  F.  is  Zx^  —  x-\-\. 
Hence  this  is  the  H.  C.  F.  of  (1)  and  (2). 

The  work  may  be  conveniently  arranged  thus  : 


(1)     9'a:*             -    a:2  +  2a:-l 

27x6            +8x2-3x  +  l 

(2) 

9a:*4-6a:8             +3a: 

27x6-3x8  +  6x2-3x 

(4)            -6x8-    x^-    x-1 

3x8  +  2x2           +1 

(3) 

6x8+4x2           +2 

3a4J_    ar2+    x 

(5)  3x2-    a-^i  3a.2_    ^  ^i     (6) 

The  object  at  each  step  is  to  obtain  a  new  expression  of  as  low  a 
degree  as  possible.  For  this  purpose  the  highest  powers  are  elimi- 
nated step  by  step  by  the  method  of  addition  or  subtraction. 

E.g.  In  Ex.  1,  x^  was  eliminated  first,  then  x^,  and  then  x8. 

By  principles  1  and  2,  each  new  expression  contains  all  the  factors 
common  to  the  given  expressions.  Hence,  whenever  an  expression  is 
reached  which  is  identical  with  the  preceding  one,  this  is  the  H.  C.  F. 


344  FACTORING 

102.    The  process  is  further  illustrated  as  follows : 

Ex.2.    FindtheH.C.F.  of  2aj3-2a;2-3a;-2, 
and  3a^-x'-2x-16. 

Arranging  the  work  as  in  Ex.  1,  we  have 

(1)     2a:8-23:2-    Sx-2  3  x^  -      x^  -    2x-    16  (2) 

2.(1)    4a;8-4a;2-    Qx-4:  6x^-   2x^-    4a;-    32  2.(2) 

x-{3)    4:X»  +  5x^-2Qx  Qx^-    Q  x^  -   9x-     6  3  •  (1) 

(4)            -9a;2+20a;~4                          4a;2+    5x~    26  (3) 

9x2- 18  a:                                36  a:2  +  45  a;  -  234  9.(3) 


4a;2 

'+    5a:- 

26 

36a:2 

!  +  45  a:  - 

234 

~36a;2 

'  +  80  X  - 

16 

(7)  2a;-4  >-36a:2  +  80x-    16      4.(4) 

(8)  a: -2  125  x- 250  (5) 

X-      2  (6) 

Explanation.  To  eliminate  x^,  we  multiply  (1)  by  3  and  (2)  by  2 
and  subtract,  obtaining  (3). 

To  eliminate  x2  from  (3),  we  need  another  expression  of  the  second 
degree.  To  obtain  this  we  multiply  (1)  by  2  and  (3)  by  x  and  sub- 
tract, obtaining  (4). 

Using  (4)  and  (3),  we  eliminate  x2,  obtaining  (5).  Since  (5)  con- 
tains all  factors  common  to  (1)  and  (2),  and  since  125  is  not  such  a 
factor,  this  is  discarded  without  affecting  the  H.  C.  F.,  giving  (6). 

Multiplying  (6)  by  9  and  adding  to  (4)  we  have  (7).  Discarding 
the  factor  2  gives  (8)  which  is  identical  with  (6).  Hence  x  —  2  is  the 
H.  C.  F.  sought. 

103.  Any  monomial  factors  should  be  removed  from  each 
expression  at  the  outset.  If  there  are  such  factors  common  to 
the  given  expressions,  these  form  a  part  of  the  H.  C.  F. 

When  this  is  done,  then  any  monomial  factor  of  any  one  of 
the  derived  expressions  may  be  at  once  discarded  without 
affecting  the  H.  C.  F.,  as  in  (5)  of  the  preceding  example. 

In  this  way  also  the  hypothesis  of  principle  2  is  always  fulfilled ; 
namely,  that  at  every  step  the  multiplier  of  one  expression  shall  have 
no  factor  in  common  with  the  other  expression. 


COMMON  FACTORS  AND  MULTIPLES  345 

Ex.  3.   Find  the  H.  C.  F.  oi  Sa^-7 x'-\-Sx-2, 
and  x*  —  a^  —  x^-~x  —  2. 

(1)  3a;8-    7x^+    dx-   2  x^  -    x^-     x^  -    x-2  (2) 

4.(1)  12a:3-28a;2  +  12a;-    8  Zx^-dx^-dx'^-Sx-Q  3.(2) 

3.(3)  12a:8-18a:2-    3  a: -18  dx* -7  x^ +  S  x^ -2  x  x-{l) 

(4)  -10a;2  +  15a;+10  4x3-6a;2-    a:  -  6  (3) 

(5)  -5(2x  +  l)(a:-2). 

Explanation.  To  eliminate  x^,  we  multiply  (1)  by  x  and  (2)  by  3 
and  subtract,  obtaining  (3). 

To  eliminate  x%  we  multiply  (1)  by  4  and  (3)  by  3  and  subtract, 
obtaining  (4). 

At  this  point  the  work  may  be  shortened  by  factoring  (4)  as  in 
(5).  We  may  now  reject,  not  only  the  factor  —  5,  but  also  2  a:  +  1, 
which  is  a  factor  of  neither  (1)  nor  (2),  since  2  a:  does  not  divide  the 
highest  power  of  either  expression.  But  a:  —  2  is  seen  to  be  a  factor 
of  (2),  by  §§  91,  92,  and  hence  it  is  a  common  factor  of  (2)  and  (4) 
and  therefore  of  (1)  and  (2).     Hence  a:  —  2  is  the  H.  C.  F.  sought. 

EXERCISES 

Find  the  H.  C.  F.  of  the  following  pairs  of  expressions : 

1.  a^+6a^-^6a+5y  a^-{-4:a^-4.a-{-5. 

2.  x^-2x^-2x^  +  5x-2y  x^-4:X^  +  6a^-5x  +  2. 

3.  2a^-9x'-lSx-4.,  a^-12a^  +  Slx-{-2S. 

4.  x^-5x^-\-Sx-2,  x*-Sa^-\-Sa^-Sx  +  2. 

5.  2a^-9ar^  +  8x-2,  2^^  ■i-5x'-5x  +  l. 

6.  3a*-2a-'  +  10a2-6a+3,  2a* +  3  a^^.  5  ^2^  g^  _3^ 

7.  15a;*  +  19a^-44a^-15a;-f-9, 

15ai'- 60^-^-510^  +  11x^15. 

8.  r^  +  2r*-27^-Sr'-7r-2,  r^-2r^-2r^-i-Ar'-\-r-2. 

104.  The  following  principle  enables  us  to  find  the  L.  CM.  of 
two  expressions  by  means  of  the  method  which  has  just  been 
used  for  finding  the  H.  C.  F. 


346  FACTORING 

The  L.  C.  M.  of  two  expressions  is  equal  to  the  product  of 
either  expression  and  the  quotient  obtained  by  dividing 
the  other  by  the  H.  C.  F.  of  the  two  expressions. 

For  let  A  and  B  be  two  expressions  whose  H.  C.  F.  is  F  so  that 
A  —  rtiF  and  B  —  nF.  Hence  the  L.  C.  M.  of  A  and  B  is  mnF.  But 
mnF  =  m(nF)  =  mB.  Also  mnF  =  n(mF)  =  nA.  Therefore  the 
L.  C.  M.  is  either  mB  or  nA,  where  m  =  A  -^  F  and  n  =  B  -i-  F. 

Ex.   Find  the  L.C.M.  of  9  a;^-a^  + 2a;- 1,  (1) 

and  27a^  +  8a;2-3a;4-l.  (2) 

The  H.  C.  F.  was  found  in  §  101  to  be  3  ^2  -  a;  +  1. 
Dividing  (1)  by  3  a;^  —  a;  +  1  we  have  3  a;^  +  x  —  1. 

Hence  the  L.  C.  M.  of  (1)  and  (2)  is 

(27a;6  +  8a;2-  3  a:  +  1)(3  x2  +  a;  -  1). 

EXERCISES 

Find  the  L.  C.  M.  of  each  of  the  following  sets. 

1.  a*  +  a^-h2a'-a  +  3,  a'  +  2w'-^2a^-a-{-^. 

2.  a^-6a^-^lla-6,a^-9a'  +  26a-24.. 

3.  2a^+3a'b-2ab^-3b%  2 a^ -  a^b - 2 a'b^ -\- 4: ab^ - 3 b\ 

4.  2a^-a^b-13ab^-6b% 

2  a^- 5  a'b -11  a'b^  +  20  aft^  + 12  b\ 

5.  4a^-15a2-5a-3,  8  a*-34:a^-i-5a'-a  +  3, 
2a«-7a2+lla-4. 

<6.   a^^a^  +  l,  a«  +  2a2-2a  +  3. 

7.  21(^-m-13kP-^5l%  3k'-16m  +  24.kr'-7P, 

8.  12  r*~  20  r's- 15  7^^"+ 35  rs^- 12  s\ 
6r3-7r2.9-~llrs2+126*3. 

9.  2a^-7a^-^6a-2,  a3+2 a^- 13 a  + 10,  «»+ 6 a^+Ga+S. 

10.   x^-xy'-{-ya^-f,  2  3^ -^  x'y  +  xy' -{- 2  f, 
2a^-{-3x'y+3xy^+2f, 


•       COMMON  FACTORS   AND   MULTIPLES  347 

PROBLEMS  INVOLVING  DENSITIES 

105.  If  a  cubic  foot  of  a  certain  kind  of  rock  weighs  2.5 
times  as  much  as  a  cubic  foot  of  fresh  water,  the  density  of  this 
rock  is  said  to  be  2.5. 

A  cubic  centimeter  of  distilled  water,  which  weighs  one  gram 
at  a  temperature  4°  above  zero,  is  used  as  a  standard  of  com- 
parison. We  therefore  say  that  the  density  of  any  substance 
is  equal  to  the  number  of  grams  which  a  cubic  centimeter  of  it 
weighs.  That  is,  the  weight  of  an  object  in  grams  is  the 
product  of  its  volume  in  cubic  centimeters  multiplied  by  its 
density. 

Hence,  if  we  represent  the  weight  of  an  object  in  grams  by  w,  its 
volume  in  cubic  centimeters  by  y,  and  its  density  by  d,  we  have  the 
relation, 

w=vd.  (1) 

1.  If  500  ccm.  of  alcohol,  density  .79,  is  mixed  with  300 
ccm.  of  distilled  water,  what  is  the  density  of  the  mixture  ? 

The  volume  of  the  mixture  is  the  sum  of  the  volumes,  and  the 
weight  of  the  mixture  is  the  sum  of  the  weights  of  the  water  and 
alcohol.     Hence,  from  formula  (1)  : 

500  X  .79  +  300  X  1  =  c?  X  800.     To  find  d. 

2.  How  many  cubic  centimeters  of  cork,  density  .24,  must 
be  combined  with  75  ccm.  of  steel,  density  7.8,  in  order  that 
the  average  density  shall  be  equal  to  that  of  water,  i.e.  so  that 
the  combined  mass  will  just  float  ? 

Let  V  =  volume  of  cork  to  be  used.  Then  the  total  volume  is 
75  +  V,  the  total  weight  is  75  x  7.8  +  .24  v,  and  the  density  is  1. 
Hence,  75  x  7.8  +  .24 1?  =  1  •  (75  +  v).     Solve  this  equation  for  v. 

3.  Brass  is  an  alloy  of  copper  and  zinc.  How  many  cubic 
centimeters  of  zinc,  density  6.86,  must  be  combined  with  100 
ccm.  of  copper,  density  8.83,  to  form  brass  whose  density  is 
8.31? 


348  FACTORING 

4.  Coinage  silver  is  an  alloy  of  copper  and  silver.  How 
many  ccm.  of  copper,  density  8.83,  must  be  added  to  10  ccm.  of 
silver,  density  10.57,  to  form  coinage  silver,  whose  density  is 
10.38? 

5.  The  density  of  pure  gold  is  19.36  and  of  nickel  8.57. 
How  many  ccm.  of  nickel  must  be  mixed  with  10  ccm.  of  pure 
gold  to  form  14  karat  gold  whose  density  is  14.88  ? 

6.  How  much  mercury,  density  13.6,  must  be  added  to  20 
ccm.  of  gold,  density  19.36,  so  that  the  density  of  the  compound 
shall  be  16.9? 

7.  What  is  the  average  density  of  40  ccm.  of  water,  den- 
sity 1,  and  180  ccm.  of  alcohol,  density  .79  ? 

8.  How  many  cubic  centimeters  of  water  must  be  mixed 
with  350  ccm.  of  alcohal,  so  that  the  density  of  the  mixture 
shall  be  97  ? 

9.  The  density  of  copper  is  8.83.  500  ccm.  of  copper  is 
mixed  with  700  ccm.  of  lead,  whose  density  is  11.35.  What  is 
the  density  of  the  combined  mass  ? 

10.  When  960  ccm.  of  iron,  density  7.3,  is  fastened  to  8400 
ccm.  of  white  pine,  the  combination  just  floats,  i.e.  has  a  density 
of  1.     What  is  the  density  of  white  pine  ? 

11.  How  many  cubic  centimeters  of  matter,  density  4.20, 
must  be  added  to  150  ccm.  of  density  8.10  so  that  the  density 
of  the  compound  shall  be  5.4  ? 

12.  How  many  cubic  centimeters  of  nitrogen,  density 
0.001255,  must  be  mixed  with  210  ccm.  of  oxygen,  density 
0.00143,  to  form  air,  whose  density  is  0.001292  ? 


CHAPTER  VI 
POWERS  AND  ROOTS 

106,  Each  of  the  operations  thus  far  studied  leads  to  a  single 
result. 

E.g.  Two  numbers  have  one  and  only  one  sum,  §  2,  and  one  and 
only  one  product,  §  7. 

When  a  number  is  subtracted  from  a  given  number,  there  is  one 
and  only  one  remainder,  §  6. 

When  a  number  is  divided  by  a  given  number,  there  is  one  and 
only  one  quotient,  §  11. 

We  are  now  to  study  an  operation  which  leads  to  more  than 
one  result ;  namely,  the  operation  of  finding  roots. 

Thus  both  3  and  —  3  are  square  roots  of  9,  since  3-3  =  9,  and  also 
(_  3)  (_  3)  =  9.     The  two  square  roots  of  9  are  written  i  V^  =  ±  3. 

The  operations  of  addition,  subtraction,  multiplication,  and 
division  are  possible  in  all  cases  except  dividing  by  zero,  which 
is  explicitly  ruled  out,  §§  24,  25. 

Division  is  possible  in  general  because  fractions  are  admitted  to 
the  number  system,  and  subtraction  is  possible  in  general  because 
negative  numbers  are  admitted.     Thus  7  -^  3  =  2J,  5  —  7  =  —  2. 

107.  The  operation  of  finding  roots  is  not  possible  in  all 
cases,  unless  other  numbers  besides  positive  and  negative  in- 
tegers and  fractions  are  admitted  to  the  number  system. 

E.g.   The  number  V2  is  not  an  integer  since  1^  =  1  and  2^  =  4. 
Suppose  V2  =  -  a  fraction  reduced  to  its  lowest  terms,  so  that  a 
and  b  have  no  common  factor.     Then  7;;  =  2.    But  this  is  impossible, 

for  if  62  exactly  divides  a^,  then  a  and  b  must  have  factors  in  com- 
mon.    Hence  V2  is  not  a,  fraction. 

349 


350  POWERS   AND   ROOTS 

108.  If  a  positive  number  is  not  the  square  of  an  integer  or 
a  fraction,  a  number  may  be  found  in  terms  of  integers  and 
fractions  whose  square  differs  from  the  given  number  by  as 
little  as  we  please.     See  p.  182,  E.  C. 

E.g.  1.41,  1.414,  1.4141  are  successive  numbers  whose  squares  differ 
by  less  and  less  from  2.  In  fact  (1.4141)^  differs  from  2  by  less  than- 
.0004,  and  by  continuing  the  process  by  which  these  numbers  are 
found,  §  176,  E.  C,  a  number  may  be  reached  whose  square  differs 
from  2  by  as  little  as  we  please. 

1.41,  1.414,  1.4141,  etc.,  are  successive  approximations  to  the 
number  which  we  call  the  square  root  of  2,  and  which  we  represent  by 
the  symbol  V2. 

109.  Definition.  If  a  number  is  neither  an  integer  nor  a 
fraction,  but  if  it  can  be  approximated  by  means  of  integers 
and  fractions  to  any  specified  degree  of  accuracy,  then  such  a 
number  is  called  an  irrational  number.     See  §  36. 

E.g.  \/2,  \/2,  V5,  etc.,  are  irrational  numbers,  whereas  Vi,  V8, 
are  rational  numbers. 

It  will  be  found  also  in  higher  work  that  there  are  other  irrational 
numbers  besides  indicated  roots.  For  instance,  the  number  tt,  which 
is  the  ratio  of  the  circumference  to  the  diameter  of  a  circle,  is  an  ir- 
rational number  though  it  is  not  an  indicated  root. 

It  is  shown  in  higher  algebra  that  irrational  numbers  corre- 
spond to  definite  points  on  the  line  of  the  number  scale,  §  48, 
E.  C,  just  as  integers  and  fractions  do. 

We,  therefore,  now  enlarge  the  number  system  to  include 
irrational  numbers  as  well  as  integers  and  fractions. 

The  set  of  numbers  consisting  of  all  rational  and  irrational 
numbers  is  called  the  real  number  system. 

110.  Even  with  the  number .  system  as  thus  enlarged,  it  is 
still  not  possible  to  find  roots  in  all  cases.  '  The  exception  is 
the  even  root  of  a  negative  number. 


THE   COMPLEX  NUMBER  351 

E.g.  V  -  4  is  neither  +  2  nor  —  2,  since  (+  2)2  =  +4  and  (-  2)^ 
=  +  4,  and  no  approximation  to  this  root  can  be  found  as  in  the  case 
of  V2. 

111.  Definition.  The  indicated  even  root  of  a  negative  num- 
ber, or  any  expression  containing  such  a  root,  is  called  an 
imaginary  number,  or  more  properly,  a  complex  number.  All 
other  numbers  are  called  real  numbers. 

E.g.  V— 4,  V— 2,  1  +  V—  2,  are  complex  numbers,  while  5,  y/2, 
1  +  V^  are  real  numbers. 

Complex  numbers  cannot  be  pictured  on  the  line  which  represents 
real  numbers,  but  another  kind  of  graphic  representation  of  complex 
numbers  is  made  in  higher  algebraic  work,  and  such  numbers  form 
the  basis  of  some  of  the  most  important  investigations  in  advanced 
mathematics. 

112.  With  the  number  system  thus  enlarged,  by  the  admis- 
sion of  irrational  and  complex  numbers,  we  have  the  following 
fundamental  definition. 

That  is,  a  fctli  root  of  any  number  n  is  such  a  number  that, 
if  it  be  raised  to  the  A;th  power,  the  result  is  n  itself. 

E.g.   (v^)8  =  2,  (\/4)2  =  4,  (V32)2  =  -2. 

The  imaginary  or  complex  unit  is  V— 1.     By  the  above 

definition  we  have  

(V-l)^  =  -l. 

In  operating  upon  complex  numbers,  they  should  first  be  ex- 
pressed in  terms  of  the  imaginary  unit. 

E.g.    V^^  =  V2  .  >/^^,  \/^=T6  =  Vl6  •  V^  =  4V^. 

V34  .  v^39  =  (V4 .  V^)(\^  V:ri)=:2  .  3(  V3T)2==  _6. 

>/z:4  +  ^fZTQ  =  Vi . ./:_  1  +.V9 .  V^^  =  (2  +  3)>/^n:=  5V^^. 

AA=n:6^  vre.  v^^^  Vi6^4 


352  POWERS  AND  ROOTS 

113.  By  means  of  irrational  and  complex  numbers  it  can  be 
shown  that  every  number  has  two  square  roots,  three  cube 
roots,  four  fourth  roots,  etc.     See  §  195,  Exs.  17-20. 

E.g.  The  square  roots  of  9  are  +  3  and  —  3.  The  square  roots 
of  -  9  are  ±  V^  =  i  3  V^^.  The  cube  roots  of  8  are  2,  -  1  +  V^ 
and  -1- V^Ts.  The  fourth  roots  of  16  are  +2,  -2,  +2  V^l 
and  -  2  V^^. 

Any  positive  real  number  has  two  real  roots  of  even  degree, 
one  positive  and  one  negative. 

E.g.    ±  2  are  fourth  roots  of  16.     The  square  roots  of  3  are  ±  V3. 

Any  real  number,  positive  or  negative,  has  one  real  root  of 
odd  degree,  whose  sign  is  the  same  as  that  of  the  number  itself. 


E.g.    V27  =  3  and  V-32  =  -  2. 

114.  The  positive  even  root  of  a  positive  real  number,  or  the 
real  odd  root  of  any  real  number,  is  called  the  principal  root. 

The  positive  square  root  of  a  negative  real  number  is  also 
sometimes  called  the  principal  imaginary  root. 

*■  E.g.  2  is  the  principal  square  root  of  4,  3  is  the  principal  4th  root 
of  81 ;  —4  is  the  principal  cube  root  of  —64;  and  +  V— 3  is  the 
principal  square  root  of  —  3. 

Unless  otherwise  stated  the  radical  sign  is  understood  to  in- 
dicate the  principal  root. 

Since  a  number  has  more  than  one  root  it  is  necessary  to 
limit  certain  theorems  so  as  to  make  them  apply  to  principal 
roots  only. 

Thus  the  square  root  of  2^  is  not  2*^  unless  the  principal  square  root 
of  2*  is  understood. 

Again,  the  cube  root  of  2^  •  3^  is  not  equal  to  2  •  3^  unless  the  principal 
cube  root  is  understood. 

Q6  Q2 

Also  the  cube  root  of  —  is  not  —  unless  it  is  specified  that  the 
48  4 

principal  cube  root  is  the  one  taken. 

Unless  this  restriction  is  understood  it  may  be  easily  shown 
that  the  conclusions  in  §§  119,  120,  121  are  not  true. 


PRINCIPLES  INVOLVING  POWERS  AND  ROOTS       353 
PRINCIPLES  INVOLVING  POWERS   AND   ROOTS 

115.  From  §  43,  (2^)2  =  {2y  =  2«  =  64. 

In  general,  if  n  and  k  are  any  positive  integers, 

For  (b^y  =  l/'  -b^'b"  ...  to  n  factors 

z=  J*+t+*.«.  ton  terms  _  Jn*^ 

Likewise,  (&")»  =  b^. 

Hence :    TJie  nth  power  of  the  kth  power  of  any  base  is  the  nkth 
power  of  that  base, 

116.  Again,  from  §§  43,  44,  (2«  •  3^)2  =  2«  •  3*. 

In  general,  if  A;,  r,  and  w  are  any  positive  integers, 

For  (a^&O"  =  (a^^O  *  (a^^O  —  to  n  factors 

=  (a*  •  a*  •••  to  n  f actors) (ft** •  J*"  •••  to  n  factors) 
=  (a^y  .  (b'^y  =  a'^^b'": 

Hence:   The  nth  power  of  the  product  of  several  factors  is  the 
product  of  the  nth  powers  of  those  factors. 

117.  From  §  43,  we  have  (1)'  =  !  '  |  =|- 
In  general,  (f.)"  =  £- 

For  we  have  {^Y  =  ^*  .  ^*  •  ^ ...  to  n  faxjtors 

W)       b'-     V     b' 

*  =^. 

Hence :    The  nth  power  of  the  quotient  of  two  numbers  equals 
the  quotient  of  the  nth  power  of  those  numbers. 


354  POWERS  AND   ROOTS 

118.  It  follows  from  §§  115, 116,  117,  that 

Any  positive  integral  power  of  a  monomial  is  found  by  mul- 
tiplying the  exponeiits  of  the  factors  by  the  expojient  of  the  poiver. 

119.  We  may  easily  verify  that  VS'  =  3^^  =  32  =  9. 

In  general,  if  k  and  r  are  positive  integers  and  b  any  posi- 
tive real  number,  we  have : 

For,  from  §  115,  (&*)'•  =  bK 

Hence  by  definition  &*  is  an  rth  root  of  h'^,  and  since  &*  is  real  and 
positive,  it  is  the  principal  rth  root  of  b^'"  (§  114). 

That  is,  </W  =  &*'•-'•  =  bK 

Hence:  77ie  rth  root  of  the  krth  power  of  any  positive 
real  number  is  a  power  of  that  number  ivhose  exponent  is 
kr-^r=k. 

E.g.  V2^  =  212^4  =  08  =  8.     But  it  does  not  follow  that 

\/(-2)12     =  (_  2)124-4  ^   (_  2)8  ::^   _  8, 

since  (-  2)i2  =(2)i2,  and  hence  ^(- 2)i-^  =  v^i2i  =  +  8. 

The  corresponding  principle  holds  when  b  is  negative  if  r  is 
odd  and  also  when  b  is  negative  if  7c  is  even. 


E.g.  V(-2)6  =(-2)6-3=  (-2)2  =4;   ^(_2)i6   =(-2)5  =  -32. 

120.   Another  general   principle,   if  a  and  b  are  any  real 
-numbers  and  r  any  positive  integer,  is 

■Vab  =  Va   •  Vi&  . 

For  by  §116,    (v^  •  Vb  y  =  (</a  y  -  (Vb  y, 

And  by  §  112,  =  a  •  &. 

Hence  ab  =  (y/a  .  Vb  y. 

Taking  the  principal  rth  root  of  both  members, 
we  have  v/a6   =  Va   .  Vb  . 


PRINCIPLES   INVOLVING  POWERS   AND   ROOTS        355 

Hence :  Tlie  nth  root  of  the  product  of  two  positive  real 
numbers  equals  the  product  of  the  nth  roots  of  the  numbers. 

When  r  is  even  the  corresponding  principle  does  not  hold  if 
a  and  h  are  both  negative. 


For  example,  it  is  not  true  that  v(  —  4)(—  9)=  V— 4-V—  9. 
For  V'(-4)(-9)  =  V36  =  6 ;  while  V^^  •  ^/^~Q 

=  2  V^Tl .  3  y/Z7i  =  6. ( V^2  =  _  6.     See  §  112. 

121.   Again,  if  a  and  h  are  any  positive  real  numbers  and  r 
is  any  positive  integer, 

'■/?  — 2^ 

*For  we  have  by  §§  117,  112,  {^\  =  ^^  -  ^ 
Hence,  taking  the  principal  rth  root  of  both  members, 


we  have 

That  is :    Tlie  rth  root  of  (lie  quotient  of  two  positive  real 
numbers  equals  the  quotient  of  the  rth  roots  of  the  numbers. 

E.g:  J16^V16^4     i/Hl=v:El  =  Zl2=_2. 

^25      V25     5'    ^  27        ^07         3  3 

The  corresponding  principle  does  not  hold  when  r  is  even 
if  a  is  positive  and  b  is  negative.     Thus  it  is  not  true  that 

2V^       2y/~^l         2'/— T- 


^-9     V=^     3V^     3(V-T)2         -3  3 

But  we  have  J-i-  =  \/l(-l)  =  2  \/^=T. 

^-9      >'9^       ^      3 

If  r  is  odd,  the  principle  holds  for  all  real  values  of  a  and  6. 


856  POWERS  AND  ROOTS 

122.   From  §§  119,  120,  121,  it  follows  that : 

If  a  monomial  is  a  perfect  power  of  the  kth  degree,  its  kth  root 
may  be  found  by  dividing  the  exponent  of  each  factor  by  the  index 
of  the  root. 

In  applying  the  above  principle  to  the  reduction  of  algebraic 
expressions  containing  letters,  it  is  assumed  that  the  values  of  the 
letters  are  such  that  the  principles  apply. 


EXERCISES 

Eind  the  following  indicated  powers  and  roots,  and  reduce 
each  expression  to  its  simplest  form : 

1.  (a^b^cfy.  4.  (a^-yy'+'y+y\  7.  (3^  •  4^  •  2^)«-*. 

2.  (2-+^ . 3*= . 5^)«-^     5.  {xyz'^+^y-y.        S,  V32«.2«.5^'.* 


-  27  .  8  a« 


64  c»d^ 

10.  (a'«+"-i6"'-"c'"")'"+".  13^   '"-y/s"'-^'  •  4"-^  •  5**'-*' . 

11.  (3«+*.4^-7.5-^)'*=^.  14.    V64.25.256.625. 

12.  ^/3*» .  42-  .  5««  .  7^  .  15.    -v/27  .  125  .  64  .  3««. 

16.    (a  -  6)'"-"(6  -  c)'^-\a  +  by-\ 


j(a-by{a'  +  2ab-{-b^ 
'   ^        (a-by(a-\-by 

18 


/(4  a.-^  +  4  a;  +  1)(4  ^  -  4  a;  +  1) 
*    \  36flJ^-12a;2^jL 

19.    -^(-  343)(-  27)a;«(a  +  b)"^  . 

20      3/(-8)(-27)(-125K-6^>^ 
-"  •     \(-  1)(_  512)(1000)a;i^Y^ 


ROOTS  OF  POLYNOMIALS  367 

ROOTS  OF  POLYNOMIALS 

123.  In  the  Elementary  Course,  pp.  178-180,  it  was  shown 
that  the  process  for  finding  the  square  root  of  a  polynomial  is 
obtained  by  studying  the  relation  of  the  expressions,  a +  6, 
a  +  6  +c,  and  a -j- b -\- c -{- d,  to  their  respective  squares. 

The  process  for  finding  the  cube  root  of  a  polynomial  is 
obtained  by  studying  the  relation  of  the  cube 

a3  +  3  a-b  +  3  aft^  +  6«  or  a'  +  6  (3  a^  +  3  a6  +  6^ 
to  its  cube  root  a  +  6. 

An  example  will  illustrate  the  process. 

Ex.  1.   Find  the  cube  root  of 

27  m^-h  108  m^n  + 144  mn^-^-  64  n\ 

Given  cube,                       27  m«+108  m2w+144  mn2+64  n^  |3  m+4  w.  cube  root 
a3  =  27  m8 1st  partial  product 


3a2=27  7H2 
3a6  =  36m« 

62=16/l2 

3  a2+3  a&+&3=27  m^+36  mw+16  n^ 


108  wi%+144  mn2+64  n»,        1st  remainder 
108  m2n+144  mn^+Gi  n«=  6(3  a^+3ab+b^ 


Explanation.  The  cube  root  of  the  first  term,  namely  3  m,  is  the 
first  term  of  the  root  and  corresponds  to  a  of  the  formula.  Cubing 
3  m  gives  27  rrfi  which  is  the  a*  of  the  formula. 

Subtracting  27  m^  leaves  108  m^  +  144  jnn^  +  64  n*,  which  is  the 
6(3  a2  +  3  a6  +  62)  of  the  formula. 

Since  6  is  not  yet  known,  we  cannot  find  completely  either  factor 
of  6(3  a2  +  3  a6  +  62),  but  since  a  has  been  found,  we  can  get  the  first 
term  of  the  factor  3  a2  +  3  rt6  +  62  ;  viz.  3  a2  or  3(3  m) 2  =  27  m%  which 
is  the  partial  divisor.  Dividing  108  m^  by  27  m^  we  have  4  n,  which 
is  the  6  of  the  formula. 

Then  3  a2  +  3  a6  +  62  =  3(3  m)*  +  3(3  wi)(4  n)  +  (4  n)2  =  27  m^ 
+  36  mn  +  16  n2  is  the  complete  divisor.  This  expression  is  then 
multiplied  by  6  =  4  n,  giving  108  mhi  +  144  mn^  +  64  n%  which  corre- 
sponds to  6(3a2  +  3a6  +  62)  of  the  formula.  On  subtracting,  the 
remainder  is  zero  and  the  process  ends.  Hence,  3  m  4-  4  n  is  the 
required  root. 


358  POWERS  AND   ROOTS 

Ex.  2.     Eind  the  cube  root  of 

33  a;^  -  9  oj^  +  a;^  -  63  a^  4-  66  a^  -  36  a;  +  8. 
We  first  arrange  the  terms  with  respect  to  the  exponents  of  x, 

jc2  _  3  X  +  2,  cube  root 

Given  cube,  7fi  —  9x^-{-33x^  —  63z^  +  mx^-mx  +  S 

a3  =  a;6 


3  a2  =  3  a;4 

3a2  +  3a6  +  62  =  3a;4-9a;8  +  9x2 

3  a'2  =  3(a;2  _  3  a;)2  =  3  x4  — 18  a;3  +  27  a;2 
3  a'2  + 3a'6'  +  6'2=3a;4- 18  a;3  + 33x2-18x4- 4 


-  9x5  +  33 x4-63x« +  66x2 -36x4-8 

—  9x5  +  27x4  —  27x8 


6x4  — 36x8  + 66x2- 36  X +  8 
6x4- 36x8  +  66x2- 36x  + 8 


^  The  cube  root  of  x%  or  x^,  is  the  first  term  of  the  root.  The  first  par- 
tial divisor,  which  corresponds  to  3  a^  of  the  formula,  is  3(a;2)2  =  SxK 
Dividing  —  9  a;5  by  3x^  we  have  —  S  x,  which  is  the  second  term  of  the 
quotient,  corresponding  to  b  of  the  formula. 

After  these  two  terms  of  the  root  have  been  found,  we  consider 
a:2  — 3  a:  as  the  a  of  the  formula  and  call  it  a'.  The  new  partial  divisor 
is  3  a'2  =  3(a;2  -  3  a;)2  =  3  x*  -  18  a;8  +  27a:2,  and  the  new  b,  which  we 
call  b',  is  then  found  to  be  2. 

Substituting  x^  -  3  x  for  a'  and  2  for  b'  in  3  a'2  +  3  a'b'  +  b'%  we 
have  3  a:4  —  18  a;8  +  33  a;2  —  18  a:  +  4,  which  is  the  complete  divisor. 
On  multiplying  this  expression  by  2  an'd  subtracting,  the  remainder 
is  zero.     Hence  the  root  is  a;2  —  3  a:  +  2. 

In  case  there  are  four  terms  in  the  root,  the  sum  of  the  first 
three,  when  found  as  above,  is  regarded  as  the  new  a,  called 
a".  The  remaining  term  is  the  new  b  and  is  called  b".  The 
process  is  then  precisely  the  same  as  in  the  preceding  step. 

EXERCISES 

Eind  the  square  roots  of  the  following : 

1.  m2  +  4mn  +  6mZ  +  4w2  +  12Zn  +  9?. 

2.  4a^  +  8aa^  +  4a2a^  +  166V  +  16a62a;  +  166*. 

3.  9  a"  -  6  a6  +  30  ac  +  6  ad  +  62  _  10  6c- 2  6d  + 25  c^ 

+  10  cd-^cP. 


ROOTS   OF  POLYNOMIALS  359 

4.  9  a' -30 ab  -8  ab'  +  25  b'+5b^-\-^. 

4 

5.  ia'x*-iaba^z-\-^a-bx^z''  +  b'x'z'-4.ab'x:^  +  Aa'^z\ 

6.  a'  -6  ab  +  10  ac  -U  ad  +  9  b^  -  30  be  -\-  42  bd  +  25  c^ 

-70c(i  +  49d2. 

7.  9  a«  -  24  a'b'  - 18  aV  +  6  a^d^  +  le  fts  ^  24  6V  -  8  ^^d^ 

+  9c^«-6c^d2  +  d*. 

-2A^w'-12y'z*-\-9z\ 
Find  the  cube  root  of  each  of  the  following : 
9.    x^-Sx'y  +  Sxy^-f  +  Sa^z-exyz  +  Syh  +  Sxz^ 

-3yz^  +  z\ 

10.  a"  +  3  a^b  -\-3 a^c-\-3 ab^  +  6  abc  +  3  ac-  +  b^  +  3  b^c 

4-36c2  +  c8. 

11.  8a;«-36ar'  +  114a;*-207ajs  +  285ar'-225a;  +  125. 

12.  27 2« -. 54 a2*  + 63 aV- 44 aV  + 21  aV- 6 a'^^  +  a^. 

13.  1  -  9  /  +  39  ?/*  -  99  /  + 156  f  - 144  i/^«  +  64  y'^, 

14.  125  aj«  -  525  a^y  +  60  x^  +  1547  ic«/  -  108  x'y''  -  1701  a^ 

-  729  y\ 

15.  64  P  -  576  P  +  2160  f  -  4320  Z«  +  4860  I'-  2916  Z^  +  729. 

16.  a^  +  0  a'b  + 15  a*b^  +  20  a^fi^  + 15  a'b^  +  6ab'  +  6«. 

17.  a»  -  9  a«6  +  36  a'b^  -  84  a«6^  4- 126  a'b'  -  126  a^ft'^  +  84  a^¥ 

-3Qa^b'  +  9ab^-b\ 

18.  a^  +  6  a^ft  -  3  ah  +  12  aft^  _  12  abc  +  3  ac^  +  8  W  - 12  bh 

-\-Qbc'-(?. 

19.  343  a«  -  441  a^6  +  777  a^b^  -  531  a%^  +  444  a^^*  _  144  ^55 

+  64  6«. 

20.  a^8  + 12  a^'  +  60  a^^  +  igQ  a^  +  240  a«  + 1 92  a^  64. 

21.  27  P  + 189  F  + 198  P  -  791  Z«  -  694  Z«  +  1701  V  -  729  Z«. 


360  POWERS  AND  ROOTS 

ROOTS  OF  NUMBERS  EXPRESSED  IN  ARABIC  FIGURES 

124.  The  cube  root  of  a  number  expressed  in  Arabic  figures, 
like  the  square  root  of  such  a  number,  may  be  found  by  the 
process  used  for  polynomials.     An  example  will  illustrate. 

Ex.  1.   Find  the  cube  root  of  389,017. 

In  order  to  decide  how  many  digits  there  are  in  the  root,  we 
observe  that  108=  looo,  1003=  1,000,000.  Hence  the  root  lies  between 
10  and  100,  that  is,  it  contains  two  digits.  Since  708  _  343,000  and 
808  _  512,000,  it  follows  that  7  is  the  largest  number  possible  in  tens* 
place.     The  work  is  arranged  as  follows : 

The  given  cube,  389  017  f  70  +  3,   cube  root. 

aS  =  708  ^  343  000    1st  partial  product. 
3a2=  3-702  =  14700 


8a5  =  3.70.3=      630 
Z.2  =  32  =         9 


3a^  + dab +  1^  =  16339 


46  017    1st  remainder. 


46  017  =  6  (3  a8  +  3  a&  +  b^). 


0 


Having  decided  as  above  that  the  a  of  the  formula  is  7  tens,  we  cube 
this  and  subtract,  obtaining  46,017  as  the  remaining  part  of  the 
power. 

The  first  partial  divisor,  3  02=14,700,  is  divided  into  46,017,  giving 
a  quotient  3,  which  is  the  6  of  the  formula.  Hence  the  first  complete 
divisor,  3  a^ -\-Sab  +  Sb%  is  15,339  and  the  product,  b(3a^  +  Sab-{-  &2), 
is  46,017.  Since  the  remainder  is  zero,  the  process  ends  and  73  is  the 
cube  root  sought. 

125.  The  cube  of  any  number  from  1  to  9  contains  one,  two, 
or  three  digits ;  the  cube  of  any  number  between  10  and  99 
contains  four,  five,  or  six  digits ;  the  cube  of  any  number 
between  100  and  999  contains  seven,  eight,  or  nine  digits,  etc. 
Hence  it  is  evident  that  if  the  digits  of  a  number  are  separated 
into  groups  of  three  figures  each,  counting  from  units'  place 
toward  the  left,  the  number  of  groups  thus  formed  is  the  same 
as  the  number  of  digits  in  the  root. 


ROOTS  OF  ARABIC   NUMBERS  361 

Ex.  2.   Find  the  cube  root  of  13,997,521. 

The  given  cube,  13  997  521  |200  +  40  +  1  =  241,  cube  root. 

a^  =  200^=    8  000000 

3  a2  =  120000 

dab=    24000 

b^  =      1600 

145600 

3  a'2  =  172800 

Ba'b'=       720 

6'2=  1 


5  997  521 


5  824  000  =  &(3a2  +  3  ab  +  b^) 


173521 


173  521 


173  521  =  b'  (3  a'2  +  3  a'b'  +  6'2). 


0 


Since  the  root  contains  three  digits,  the  first  one  is  the  cube  root 
of  8,  the  largest  integral  cube  in  13. 

The  first  partial  divisor,  3  •  200^  =  120,000,  is  completed  by  adding 
3  a6  =  3  .  200  .  40  =  24,000,  and  b^  =  1600. 

The  second  partial  divisor,  3  a'^,  which  stands  for  3(200  +  40)^ 
=  172,800,  is  completed  by  adding  3  a'b'  which  stands  for  3  •  240  •  1  = 
720,  and  fe'^'^yi^jch  stands  for  1,  where  a'  represents  the  part  of  the 
root  already  found  and  b'  the  next  digit  to  be  found.  At  this  step 
the  remainder  is  zero  and  the  root  sought  is  241. 

EXERCISES 

Find  the  square  root  of  each  of  the  following: 

1.   58,081.  2.   795,664.  3.   11,641,744 

Find  the  cube  root  of  e&,ch  of  the  following : 

4.  110,592.  7.   205,379.  10.   2,146,689. 

5.  571,787.  8.   31,855,013.  11.   19,902,511. 

6.  7,301,384.  9.   5,929,741.  12.   817,400,375. 

126.  Since  the  cube  of  a  decimal  fraction  has  three  times  as 
many  places  as  the  given  decimal,  it  is  evident  that  the  cube 
root  of  a  decimal  fraction  contains  one  decimal  place  for  every 
three  in  the  cube.  Hence  for  the  purpose  of  determining  the 
places  in  the  root,  the  decimal  part  of  a  cube  should  be  divided 
into  groups  of  three  digits  each,  counting  from  the  decimal 
point  toward  the  right. 


362 


POWERS  AND   ROOTS 


Ex.   Approximate  the  cube  root  of  34.567  to  two  places  of 
decimals. 


a3  =  33  = 
3  a2  =  3  .  32  =  27. 
3a&  =  3.3(.2)=    1.8 

&'^=  (.2)2= M 

28.84 

3  a'2  =  3(3.2)2  =30.72 

3a'6'  =  3(3.2)(.05)=      .48 

&'2=(.05)2=      .0025 
31.2025 
3a"2=3(3.25)2=31.6875 
3«"6"=3(3.25)(.007)=    .06825 
6"2=(.007)2=_^000049 


34.567  [3+ .2 +  .05 +  .007  =  3.257 
27.000 


31.755799 


7.567 


5.768 


=  6(3a2  +  3a6  +  62) 


1.799000 


1.560125        =  &'(3  a'»  +  3  a'b'  +  6'2) 


.238875000 


.222290593  =  6"(3a"2  +  3  a"b" -\-  b"^) 


.016584407 
The  decimal  points  are  handled  exactly  as  in  arithmetic  work. 

Evidently  the  above  process  can  be  carried  on  indefinitely. 
3.257  is  an  approximation  to  the  cube  root  of  34.567.  In 
fact  the  cube  of  3.257  differs  from  34.567  by  less  than  the 
small  fraction  .017.  The  nearest  approximation  using  two 
decimal  places  is  3.26.  If  the  third  decimal  place  were  any 
digit  less  than  5,  then  3.25  would  be  the  nearest  approximation 
using  two  decimal  places.  Hence  three  places  must  be  found 
in  order  to  be  sure  of  the  nearest  approximation  to  two  places. 

EXERCISES 

Approximate  the  cube  root  of  each  of  the  following  to  two 
places  of  decimals. 

1.  21.4736.  6.   .003. 

2.  6.5428.  7.    .3917. 

3.  58.  8.   .5. 

4.  2.     '  9.    .05. 

5.  3.  10.    6410.37. 
16.   Approximate  the  square  root  in  Exs.  1,  2,  10,  11,  and 

15  of  the  above  list. 


11.  .004178. 

12.  200.002. 

13.  572.274. 

14.  31.7246. 

15.  54913.416. 


COMMON  FACTORS   AND   MULTIPLES  363 


PROBLEMS  ON  MOMENTUM 

127.  The  force  with  which  a  moving  body  strikes  another 
depends  both  upon  its  mass  and  upon  its  rate  of  motion.  The 
product  of  the  mass  and  velocity  of  a  moving  body  is  called 
its  momentum.  The  mass  of  a  body  is  proportional  to  its  weight. 
Hence  weight  is  often  used  instead  of  mass. 

By  careful  experiment  it  has  been  found  that  when  a  moving 
body  strikes  a  body  at  rest  but  free  to  move,  the  two  will  move 
on  with  a  combined  momentum  equal  to  the  momentum  of  the 
first  body  before  the  impact. 

Thus,  if  a  freight  car,  weighing  25  tons  and  moving  at  the  rate  of 
12  miles  per  hour,  strikes  a  standing  car  weighing  15  tons,  the  two  will 
move  on  with  the  original  momentum  of  12  •  25.  But  as  the  combined 
weight  is  now  25  +  15,  the  rate  of  motion  has  been  decreased  to  7^ 
miles  per  hour,  since  12  •  25  =  7  J  (25  +  15). 

Again,  if  a  croquet  ball  weighing  8  ounces  and  moving  20  feet  per 
second,  strikes  another  weighing  7  ounces  and  starts  it  off  at  the  rate 
of  18  feet  per  second,  then  if  the  diminished  velocity  of  the  first 
ball  is  called  y,  we  have 

8.20  =  7. 18  + Sv, 
and  solving,  v  =  4.25. 

This  indicates  that  the  first  ball  is  nearly  stopped,  which  coincides 
with  common  observation. 

1.  In  a  switch  yard  a  car  weighing  40  tons  and  moving  8 
miles  per  hour  strikes  a  standing  car  weighing  24  tons.  What 
is  the  velocity  of  the  two  after  impact  ? 

2.  A  billiard  ball  weighing  6  ounces  and  moving  16  feet 
per  second  strikes  another  ball  which  it  sends  off  at  the  rate  of 
10  feet  per  second.  The  rate  of  the  first  ball  is  reduced  to  9 
feet  per  second  by  the  impact.  What  is  the  weight  of  the 
second  ball  ? 

3.  A  bowler  uses  a  16-ounce  ball  to  take  down  the  last  pin. 
The  ball  sends  the  pin  off  at  a  velocity  of  6  feet  per  second, 


864  POWERS  AND  ROOTS 

the  weight  of  the  pin  being  48  ounces,  while  the  velocity  of  the 
ball  is  reduced  to  4  feet  per  second.  With  what  velocity  did 
the  ball  strike  the  pin  ? 

4.  In  each  of  these  problems  we  have  considered  the  weight 
of  two  bodies,  which  we  may  call  w^  and  Wg.  If  we  call  Vj  the 
velocity  with  which  the  first  strikes  the  second,  t'g  the  velocity 
imparted  to  the  second,  and  v-^  the  resulting  decreased  velocity 
of  the  first,  we  have 

IVl/l=IVl/l'  +  1^2/2.  (1) 

5.  Solve  the  equation  (1)  for  Vi  in  terms  of  Wj,  W2,  v^,  and 
%     Translate  into  words. 

6.  Solve  the  equation  (1)  above  for  w^  in  terms  of  v^,  v/,  w^, 
and  V2.     Translate  into  words. 

7.  Solve  the  equation  (1)  above  for  %'  in  terms  of  Wi,  Wg* 
Vj,  and  v^.     Translate  into  words. ' 

8.  Solve  equation  (1)  for  W2  in  terms  of  Wi,  t\,  Vi',  V2,  and 
translate  the  result  into  words. 

9.  Solve  equation  (1)  for  V2,  in  terms  of  Wj,  Wg,  v^,  Vi,  and 
translate  the  result  into  words. 

10.  In  the  result  of  the  last  exercise  substitute  Wi=i50, 
2^2  =  40,  Vi  =  10,  Vi=2f  and  find  the  value  of  Vg-  Make  a 
problem  to  fit  this  case. 

11.  In  the  result  of  problem  8  substitute  Wi  =  1000, 
Vi  =  75,  Vi  =  25,  V2  =  50,  and  find  the  value  of  Wg.  Make  a 
problem  to  fit  this  case. 

12.  In  the  result  of  problem  6  substitute  Wi  =  60,  ?;i'  =  10, 
W2  =  ^0,  '^2  =  25,  and  find  the  value  of  Wi.  Make  a  problem 
to  fit  this  case. 

13.  In  the  result  of  problem  7  substitute  ?/7i  =  250,  ^2  =  125, 
Vi  =  50,  and  V2  =  50,  and  find  the  value  of  v/.  Make  a  prob- 
lem to  fit  this  case. 


CHAPTER  VII 


QUADRATIC  EQUATIONS 
EXPOSITION  BY   MEANS   OF   GRAPHS 

128.  We  saw,  §  65j  that  a  single  equation  in  two  variables 
is  satisfied  by  indefinitely  many  pairs  of  numbers.  If  such  an 
equation  is  of  the  first  degree  in  the  two  variables,  the  graph  is 
in  every  case  a  straight  line. 

We  are  now  to  consider  graphs  of  equations  of  the  second 
degree  in  two  variables.     See  §  (dQ>. 

Ex.  1.    Graph  the  equation  y  =  ar'. 

By  giving  various  values  to  x  and  computing  the  corresponding 
values  of  y,  we  find  pairs  of  numbers  as  follows  which  satisfy  this 
equation  : 

ly  =  0.  U  =  l.   U  =  l.       U  =  4.   U  =  4.       ly  =  9.  U  =  9. 

These  pairs  of  numbers  correspond  to  points  which  lie  on  a  curve 
as  shown  in  Figure  3.     (Use  two  squares  for  one  unit  on  avaxis.) 

By  referring  to  the  graph  the 
curve  is  seen  to  be  symjuetrical 
with  respect  to  the  y-axis.  This 
can  be  seen  directly  from  the 
equation  itself  since  x  is  involved 
only  as  a  square  and  hence,  if  y  =  a;^ 
is  satisfied  hy  x  =  a,  y  =  h,\t  must 
also  be  satisfied  bya:  =  —  a,?/  =  &. 

It  may  easily  be  verified  that 
no  three  points  of  this  curve 
lie  on  a  straight  line.  The 
curve  is  called  a  parabola. 


- 

\ 

(-3 

9) 

+9 

(3 

,9) 

1 

V 

+8 

/ 

\ 

•?? 

+7 

/ 

\ 

^ 

+6 

9a 

+6 

/ 

(-2 

i>\ 

+4 

(<.» 

4) 

\ 

+3 

J 

\ 

+2 

/ 

\ 

(-1 

1) 

+1 

/o 

1) 

_ 

% 

\ 

^ 

A 

•+ 

i 

4 

3 

X 

•aa 

is 

(( 

,0) 

-I 

_J 

365 


Fia.  3. 


366 


QUADRATIC   EQUATIONS 


' 

V 

+5 

/ 

IH 

.5) 

+i 

(3 

5)/ 

\ 

+3 

/ 

\ 

•s 

+2 

i 

V 

e 

+1 

/ 

_ 

\ 

_ 

I 

1 
-1 

5ft 

0 

/ 

-f 

2 

(- 

f.O) 

V 

X- 

XX 

s 

-1 

/ 

(1,0 

) 

\ 

-2^ 

/ 

\ 

/ 

<s 

[/ 

(0, 

3) 

u, 

4) 

-4 

Fig.  4. 


Ex.  2.    Graph  the  equation 

Each  of  the  following  pairs  of 
numbers  satisfies  the  equation  : 


x  =  0. 


I' 


1^ 


I'l 


-1, 


3.   ly  =  0.   [7/ =  -4. 


2,  /a:  =  -2,    fa:  =  -3, 
5.    12/ =-3.  ly  =  0. 


U  =  5. 


Plotting  these  points  and  draw- 
ing a  smooth  curve  through  them, 


we  have  the  graph  of  the  equation,  as  in  Figure  4. 


EXERCISES 

In  this  manner  graph  each  of  the  following : 

1.  y  =  o^-l.  7.   2/  =  5iC-a^-4. 

2.  2/  =  a^  +  4a;.  8.   ?/  =  4aj  — a;^  +  5. 

3.  2/  =  a:2_|_3^_4  2.  y  =  x^ -{-5x  —  Q, 

4.  ?/  =  aj^  +  5a;  +  4.  10.  y  =  —  x^  -\-x. 

5.  2/  =  a;2-7aj  +  6.  11.  2/  =  4a^-3a;- 1. 
.6.  2/  =  3a^  —  7x  +  2.  12.  2/  =  —  4a^  +  3a;  +  l. 

129.  We  now  seek  to  find  the  points  at  which  each  of  the 
above  curves  cuts  the  a>axis.  The  value  of  y  for  all  points  on 
the  a-axis  is  zero.  Hence  we  put  y  =  Of  and  try  to  solve  the 
resulting  equation. 

Thus  in  Ex.  2  above,  if  ?/  =  0,  a;^  +  2  a:  -  3  =  (a;  4-  3)  (a:  -  1)  =  0, 
which  is  satisfied  by  a:  =  1  and  a:=:  —  3.  Hence  this  curve  cuts  the  a;-axis 
in  the  two  points  a:  =  l,  y=0  and  a:=  —  3,  y  =  0,  as  shown  in  Figure  4. 

Similarly  in  Ex.  1,  if  ?/  =  0,  x^  =  0,  and  hence  a:  =  0.  Hence  the 
curve  meets  the  ar-axis  in  the  point  a;  =  0,  ?/  =  0,  as  shown  in  Figure 
3.     On  this  point  see  §  131,  Ex.  2. 


ALGEBRAIC   SOLUTION  367 

EXERCISES 

Find  the  points  in  which  each  of  the  twelve  curves  in  the 
preceding  list  cuts  the  a^axis. 

Notice  that  in  every  case  the  expression  to  the  right  of  the  equality 
sign  can  be  factored,  so  that  when  y  =  0  the  resulting  equation  in  x 
may  be  solved  as  in  §  94. 

Ex.  3.  Plot  the  curve  y  =  a^-{-4:X  +  2  and  find  its  intersec- 
tion points  with  the  a;-axis. 

We  are  not  able  to  factor  x^-{-ix+2hj  inspection.  Hence  we  solve 
the  equation  x2  +  4a:  +  2  =  0  by  completing  the  square  as  in  §  195, 
E.  C,  obtaining  a:  =  —  2  +  V2  and  x  =  —  2  —  V2.  Hence  the  curve 
cuts  the  X-axis  in  points  whose  abscissas  are  given  by  these  values  of  x. 

In  making  this  graph,  we  first  plot  points  corresponding  to  integral 
values  of  x,  as  before;  then,  in  drawing  the  smooth  curve  through 
these,  the  intersections  made  with  the  a;-axis  are  approximately  the 
points  on  the  number  scale  corresponding  to  the  incommensurable 
numbers,  -  2  +  V^  and  -  2  -  V2.     See  §  109. 

EXERCISES 

In  this  manner,  find  the  points  at  which  each  of  the  follow- 
ing curves  cuts  the  ic-axis,  and  plot  the  curves.  For  reduction 
of  the  results  to  simplest  forms,  see  §§  168,  169,  E.  C. 

1.  2/  =  ar^  +  5a;-H3.  5.  2/  =  2a;-5a;2  +  8. 

2.  y  =  3x^  +  Sx-2,  6.  y  =  5  +  Sx-3aF. 

3.  2/=6aj-4a^-f 5.  7.  y  =  3-9a^-llx. 

4.  y  =  -4.-2x-{-5x^.  8.  y  =  -2-2x-{-3^, 

130.  Each  of  the  foregoing  exercises  involves  the  solution  of 
an  equation  of  the  general  form  ax^  -hbx  +  c  =  0.  Obviously, 
by  solving  this  equation,  we  shall  obtain  a  formula  by  means 
of  which  every  equation  of  this  type  may  be  solved.  See 
§  199,  E.  C. 

The  two  values  of  x  are : 


'^~  2^  '  '''-  2l 


868  QUADRATIC  EQUATIONS 

EXERCISES 

By  means  of  this  formula,  find  the  solutions  of  each  of  the 

following  equations : 

1.  2j»2-3a;-4=0.  11.  3x-9a^  +  l  =  Q, 

2.  3a;2H-2a;-l  =  0.  12.  7aj2~3a;-2  =  0. 

3.  3x2_2aj_l  =  0.  13.  6a^  +  7fl5  +  l  =  0. 

4.  4a;2  +  6a?  +  l  =  0.  14.  4:X^-{-5x-S  =  0. 

5.  a;2-7x  +  12  =  0.  15.  4.0^-5x^3=0, 

6.  5ic24-8a;  +  3  =  0.  16.  8ar'  +  3a;-5  =  0. 

7.  5a;2-8a;  +  3  =  0.  17.  7a^  +  a;-3  =  0. 

8.  5a^  +  8a;~3  =  0.  18.  7a^-a;-4=:0. 

9.  5a^-8a;-3  =  0.  19.  x^ -2ax  =  Sb-a\ 
10.  2a;-3a;2  +  7  =  0.  20.  af-6ax  =  A9c^-9a^ 

22.  -2c^-i^x-2c^^  =  5!i£^. 

2  2 

23.  a;2-  — +  2mn  =  47iaj. 

24.  a;2-2aa;  +  4a6  =  &2-|.3a*. 

25.  aj^  —  a^a?  +  a^h  —  ax  =  al)^  —  bx. 
2Q.  x^  +  9  —  c  =  6x. 

27.  nx^  +  mPn  =  m?i^a;  +  mic. 

28.  2(a  +  l)a^-(a-hiyx-h2(a-\-l)=:^x. 

29.  ic24-9c(^  +  3c  =  (3c  +  3d  +  l)a;. 

30.  ar^  +  2a2  +  3a-2=(3a  +  l)a;. 

131.    We  now  consider  the  intersections  of   other  straight 
lines  besides  the  (B-axis  with  curves  like  those  plotted  above. 


DISTINCT,  COINCIDENT,  AND  IMAGINARY  ROOTS    369 

Ex.  1.  Graph  on  the  same  axes  the  straight  line,  y  =  —  2 
and  the  curve,  y  =  a^-\-2x  —  3. 

This  line  is  parallel  to  the  a:-axis  and  two  units  below  it.  It  cuts 
the  curve  in  the  two  points  whose  abscissas  are  ar^  =  —  1  +  V2  and 
X2=  —  1  —  V'2,  as  found  by  substituting  —2ior7/my  =  x^+2x  —  3 
and  solving  the  resulting  equation  in  x  by  the  formula,  §  130. 

Ex.  2.   Graph  on  the  same  axes  y=  —4:  and  y  =  x^-\-2x  —  3. 
This  line  seems  not  to  cut  the  curve  but  to  touch  it  at  the  point  whose 
abscissa  is  a:  =  —  1. 

Substituting  and  solving  as  before,  we  find, 


_Q-}.  v/r-^_  -2  +  0 


=  -1 


_o  _  V4^^      -2-0  1 

and  X2= = =-1. 

In  this  case  the  two  values  of  x  are  equal,  and  there  is  only  one  point 
common  to  the  line  and  the  curve.  This  is  understood  by  thinking  of 
the  line  y=  —2,m  the  preceding  example,  as  moved  down  to  the  position 
y  =  —  4,  whereupon  the  two  values  of  x  which  were  distinct  now  coincide. 


132.  From  the  formula,  x  =  -^^y      ^^^  it  is  clear  that 

the  general  equation,  ax^-{-bx  +  c  =  0  has  two  distinct  solutions 
unless  the  expression  6^  —  4  ac  reduces  to  zero,  in  which  case  the 

two  values   of    x    coincide,   giving   x^  =  ~  ^        =  —  -—  and 
^-6-0^ 6_  2a  2a 

^'         2  a  2  a 

Ex.  1.    In    2  ar^  —  9  ic  +  8  =  0,    determine    without    solving 
whether  the  two  values  of   x  are  distinct  or  coincident. 

In  this  case,  a  =  2,  6  =  —  9,  c  =  8. 

Hence  &2  _  4  ^^  =  81  -  64  =  17. 

Hence  the  values  of  x  are  distinct. 

Ex.  2.   In  4  aj^  — 12  «  +  9  =  0,  determine  whether  the  values 
of  X  are  distinct  or  coincident. 

In  this  ca^e,  ft^  _  4  ^c  =  144  —  4.4-9  =  0.     Hence  the  values  of  x 
are  coincident. 


370  QUADRATIC   EQUATIONS 

EXERCISES 

In  each  of  the  following,  determine  without  solving  whethei 
the  two  solutions  are  distinct  or  coincident : 

1.  X--7  x  +  4:  =  0.  6.   6a^-3x-l  =  0. 

2.  4a;2  +  28a;  +  49  =  0.  7.    A  x" -16 x +  16  =  0, 

3.  9a;2  +  12.T  +  4  =  0.  8.   8a^-13  =  4a;. 

4.  a^  +  6aj  +  9  =  0.  9.    12 a^ -  18  =  24 a?. 

5.  _a;2  +  9a;  +  25  =  0.  10.    16  a^- 56  a;  = -49. 

133.  Definition.     A  line  which  cuts  a  curve  in  two  coincident 
points  is  said  to  be  tangent  to  the  curve. 

134.  Problem.     What  is  the  value  of  am  y  =  a,  if  this  line 
is  tangent  to  the  curve  y  =  o(^  +  5x  +  S? 

Substituting  a  for  y  and  solving  by  means  of  the  formula,  we  have 


^^  _5.|.  V25^r4(8-a) 
2  * 

If  the  line  is  to  be  tangent  to  the  curve,  then  the  expression  under  the 
radical  sign  must  be  zero  so  that  the  two  values  of  x  may  coincide. 
That  is,  25  -  4(8  -  a)  =  0,  or  a  =  |. 

On  plotting  the  curve,  the  line  y  =  I  is  found  to  be  tangent  to  it. 

EXERCISES 

In  the  first  18  exercises  on  p.  368  obtain  equations  of  curves 
by  letting  the  left  members  equal  y.  Then  find  the  equations 
of  straight  lines,  y  =  a,  which  are  tangent  to  these  curves. 

135.  Problem.  Find  the  intersection  points  of  the  curve 
y  =  x^-{-Sx-{-5  and  the  line  y  =  2\. 

Substituting  for  y  and  solving  for  x  we  have 


_6+V36-40      -6  +  2  V-1      -3+v 
'^^~               4               '              4              "            2 

3 

^       _6-V36-40      _6-2V-l      -3-V 

-1 

'''^-4                             4                          2 

DISTINCT,  COINCIDENT,  AND   IMAGINARY   ROOTS     371 

These  results  involve  the  imaginary  unit  already  noticed  in 
§  112.  Numbers  of  the  type  a  +  6 V—  1  are  discussed  further 
in  §  195.  For  the  present  we  will  regard  such  results  as  merely 
indicating  that  the  conditions  stated  by  the  equations  cannot 
be  fulfilled  by  real  numbers.  This  means  that  the  curve  and 
the  line  have  no  point  in  common^  as  is  evident  on  constructing 
the  graphs. 

By  proceeding  as  in  §  134  we  find  that  the  line  y  =  V^  is  tangent 
to  the  curve  y  =  a:^  +  3  a:  +  5.  Clearly  all  lines  y  =:  «,  in  which  «  >  -y^, 
are  above  this  line  and  hence  cut  this  curve  in  two  points. 

All  such  lines  for  which  a  <  ^^  are  below  the  line  y  =  J^  and  hence 
do  not  meet  the  curve  at  all. 

Solving  y  =  a  and  y  =  x^  +  3  ar  +  5  for  x  by  first  substituting  a  for  y 

we  have  ^ 

^^-3j-V4a_ll 

2 

If  a  >  J^  the  number  under  the  radical  sign  is  positive,  and  there  are 
two  real  and  distinct  values  of  x.  Hence  the  line  and  the  curve  meet 
in  two  points. 

If  a<^,  the  number  under  the  radical  sign  is  negative.  Con- 
sequently the  values  of  x  are  imaginary  and  the  line  and  the  curve  do 
not  meet. 

Hence  we  see  that  the  conchisions  obtained  from  the  solution  of 
the  equations  agree  with  those  obtained  from  tlie  graphs. 

136.  From  the  two  preceding  problems  it  appears  that  it  is 
possible  to  determine  the  relative  positions  of  the  line  and  the 
curve  without  completely  solving  the  equations.  Namely,  as 
soon  as  y  is  eliminated  and  the  equation  in  «  is  reduced  to 
the  form  axP  +  bx  -j-  c  =  0,  we  examine  b^  —  4:aG  as  follows  : 

(1)  If  6^  — 4  ac>  0,  i.e.  positive,  then  the  line  cuts  the  curve 
in  two  distinct  points. 

(2)  If  6^  _  4  ac  =  0,  then  the  line  is  tangent  to  the  curve. 
See  §  133. 

(3)  If  6^  —  4  ac  <  0,  i.e.  negative,  then  the  line  does  not  cut 
the  curve. 


372  QUADRATIC  EQUATIONS 

137.   Problem.     Find  the  points  of  intersection  of 

y=x^-^3x-{-lS  (1),   and  y-{-3x  =  7  (2). 

Eliminating  y  and  reducing  the  resulting  equation  in  a:  to  the  form 
ax^-\-  bx  +  c  =  0,  we  have  x'^  +  Qx  +  Q  =  0.     ^ 

Solving,  xi  =  -  3  +  V3,  Xg  =  -  3  -  Vs. 

Substituting  these  values  of  x  in  (2)  and  solving  for  y,  we  have 

f  X,  =  -  3  +  V3  ,     (  X2=  -S-VS 

i  -     and    <  r- 

1  yi  =  16  -  3  V3  1  ?/2  =  16  +  3  V3 

which  are  the  points  in  which  the  line  meets  the  curve. 

Here  W-  —  ^iac  =  12,  which  shows  in  advance  that  there  are  two 
points  of  intersection. 

EXERCISES 

In  each  of  the  following  determine  without  graphing  whether 
or  not  the  line  meets  the  curve,  and  in  case  it  does,  find  the 
intersection  points : 

(y  =  2x'-Sx-4.,  (y  =  5x'-{-Sx  +  3, 

\y-x  =  3,  "    [2y-5x-2  =  0. 

r    y  =  2x'  +  2x-l,  7      (y  =  5x^-Sx  +  3, 

'     [2y==x-l.  '     [3-x  =  3y. 


3. 


4. 


(y  =  3x^-2x-l,       '  g      r2/  =  -5a;2  +  8iK-3, 

\2x-y  =  4..  '     \2-4.y-x  =  ^. 


2/4-5.  [51/  — 3a;  =  8. 

g      \y^x^-lx  +  12,  ^^     ly  =  3x-3x'^  +  7, 

{5x-y  =  -l.  '    \-5-3x  +  2y==0. 

138.   Problem.   Graph  the  equation  y?-{-y^  =  25. 

Writing  the  equation  in  the  form  y  =± V25  —  a;^,  and  assigning 
values  to  a;,  we  compute  the  corresponding  values  of  y  as  follows : 

ra;  =  0,         fa:  =±5,      fa:  =  3,         fa:  =  -3,      fa:  =  4,  fa:  =  -4, 

l2/=±5.     U  =  0.         \.y=±^.     l2/=±4.     \y  =  ±Z.     {^^  =  ±3. 


DISTINCT,   COINCIDENT,   AND   IMAGINAR^  ROOTS     373 


Evidently,  for  x  greater  than  5  in  absolute  value,  the  corresponding 
y's  are  imaginary,  and  for  each  x  between  —  5  and  +  5  there  are  two 
y's  equal  in  absolute  value,  but 
with  opposite  signs. 

It  seems  apparent  that  these 
points  lie  on  the  circumference 
of  a  circle  whose  radius  is  5,  as 
shown  in  Figure  5.  Indeed,  if  we 
consider  any  point  x-^,  y^  on  this 
circumference,  it  is  evident  that 
x^  4-  y^  =  25,  since  the  sum  of 
the  squares  on  the  sides  of  a  right 
triangle  is  equal  to  the  square 
on  the  hypotenuse.  (See  figure, 
p.  173,  E.  C.) 

The  equation  x^  +  y^  =  25  is, 
therefore,    the    equation    of    a 
circle  with  radius  5.     Similarly,  x"^  -\-  y^  =  r^  is  the  equation  of  a 
circle  with  center  at  the  point  (0,  0)  and  radius  r. 

139.   Problem.   Find  the  points  of  intersection  of  the  circle 

rc^ -f- y2  __ 25  and  the  line  x-\-y  =  T. 

Eliminating  y  from  these  equations,  and  reducing  the  equation  in 
X  to  the  form  ax^  +  &a:  +  c  =  0,  we  have 


1 

1 — 

(0,5) 

t3,4) 

^ 

^ 

s, 

(3,4) 

/ 

•2 

Sa 

!) 

/ 

%-4,3) 

^ 

^ 

)k 

L 

/ 

^ 

-rt 

iU 

(0,0) 

r^ 

^ 

a-, 

"'1' 

(U)     1 

X- 

ax 

i? 

V 

\ 

(- 

,^y 

^ 

J 

(*.f> 

(-3 

HI) 

^ 

' — 

^■ 

y 

(3,^) 

5) 

—J 

Fig.  5. 


From  which 


a:2  _  7  a:  +  12 
ar,  =  4,  Xo 


Substituting  these  values  of  a:  in  a:  +  y  =  7,  we  have  y^  =  3,  yg  =  ^• 
Hence  a;j  =  4,  y^  =  3  and  ar2  =  3,  2/3  =  ^  ^^"^  ^^^  required  points. 
Verify  this  by  graphing  the  two  equations  on  the  same  axes. 

140.   Problem.   Find  the  points  of  intersection  of  the  circle 

ic2  4-  2/2  =  25  and  the  line  3  a;  +  4  y  =  25. 

ft  ,  f^ 
Eliminating  y  and  solving  for  x,  we  find  x  =     ^     =  3. 

Hence  x^  —  x.^  —  3,  from  which  y^r:::  y^z=  4. 

Since  the  two  values  of  x  coincide,  and  likewise  the  two  values  of 
y,  the  circumference  and  the  line  have  but  one  point  in  common. 
Verify  by  graphing  the  line  and  the  circle  on  the  same  axes. 


374  '       QUADRATIC  EQUATIONS 

141.   Problem.    Find  the  points  of  intersection  of 

a^-h  2/2  =  25 
and  a;  +  2/  =  10. 

Substituting  for  y  and  solving  for  x  we  have 

_  20  j:  V400  -  600  _  20  j=  V-  200 
^-  -4  -  4 

^20  j:10V^^^10tb5v/^r-2 
4  2  * 

The  imaginary  values  of  x  indicate  that  there  is  no  intersection 
■point.     Verify  by  plotting. 

EXERCISES 

In  each  of  the  following  determine  by  solving  whether 
the  line  and  the  circumference  meet,  and  in  case  they  do, 
find  the  points  of  intersection.  Verify  each  by  constructing 
the  graph. 

^^     ra^4-/=16,        ^     [^  +  y'  =  l.  9,    |a^  +  2/^  =  12, 

*[ic  +  2/  =  4.  *\aj  +  2/  =  8.  '     {x  —  y  =  Q. 

^      (x'j^f  =  ^Q,         ^      p  +  2/^  =  8,  ^^      1^  +  2/^  =  4, 


3. 


ra;2  +  /  =  25,  ra^-}.2/2  =  41,        ^^      (o?  +  f  =  m, 

{2x-\-y  =  -b.        '     [x-3y  =  7.  '     [x  +  2y  =  10. 

^      (x'  +  f  =  20,        g      (x'-^f  =  29,        ^2      (x^  +  f  =  25, 
l2a;H-2/  =  0.  '     [3x-7y=-29.       '     [x-\-y  =  9. 

142.  Problem.  Graph  on  the  same  axes  the  circle,  a:^-\-y'^=5% 
and  the  lines,  3  aj  +  4  ?/  =  20,  3  a;  +  4  2/  =  25,  and  3  a;  +  4  ?/  =  30. 

The  first  line  cuts  the  circumference  in  two  distinct  points,  the 
second  seems  to  be  tangent  to  it,  and  the  third  does  not  meet  it.  Ob- 
serve that  the  three  lines  are  parallel.     See  Figure  6. 


DISTINCT,   COINCIDENT,   AND  IMAGINARY  ROOTS    375 


In  order  to  discuss  the 
relative  ijositions  of  such 
straight  lines  and  the  circum- 
ference of  a  circle,  we  solve 
the  following  equations  simul- 
taneously : 

a^+    f  =  r^  (1) 

3a;  +  4y  =  c  (2) 

Eliminating  y  by  substitution, 
and  solving  for  x,  we  find 


H^ 

[S 

X 

1 

fOfi 

^ 

^s  V*  . 

^ 

'^ 

fcrs 

U 

X 

- 

/ 

^> 

k'<«^-> 

/ 

•2 

-^vs. 

^ 

s 

fN 

B 

\ 

N 

-5,0) 

;* 

(0,0) 

(5,0) 

s 

X 

axis 

\ 

\ 

/ 

"V 

^ 

. 

.*-■ 

y 

(0.-6) 

_ 

3  c  jr  4  V25  r'^  -  c^ 
25 


(3) 


Fig.  6. 


The  two  values  of  x  from  (3)  are  the  abscissas  of  the  points  of 
intersection  of  the  circumference  (1)  and  the  line  (2). 

These  values  of  x  are  real  and  distinct  if  25  r^  —  c^  is  positive,  real 
and  coincident  if  25  r^  —  c^  =  0,  and  imaginary  if  25  r^  —  c^  is  negative. 

Now  25  r^  —  c^  is  positive  if  r  =  5,  c  =  20;  zero  if  r  =  5,  c  =  25 ;  and 
negative  if  r  =  5,  c  =  30. 

Hence  these  results  obtained  from  the  solution  of  the  equations 
agree  with  the  facts  observed  in  the  graphs  above. 

143.  Definition.  Letters  such  as  c  and  r  in  the  above  solution 
to  which  any  arbitrary  constant  values  may  be  assigned  are 
called  parameters,  while  x  and  y  are  the  unknowns  of  the 
equations. 

EXERCISES 

Solve  each  of  the  following  pairs  of  equations. 

Give  such  values  to  the  parameters  involved  that  the  line  (a)  may 
cut  the  curve  in  two  distinct  points,  (6)  may  be  tangent  to  the  curve, 
(c)  shaU  fail  to  meet  the  curve. 


\aa;  +  32/  =  16. 


l2a;  +  6?/  = 


4. 


ra^  +  /  =  25, 

\2x-\-Sy=:c. 

f  =  Sx, 

3  a;  4-  4  2/  =  c. 


376 


QUADRATIC  EQUATIONS 


5. 


7. 


8. 


y  =  ocF-^  mx  -\-  4, 

y  =  L 
y  =  ma?  —  ic  —  4, 
8. 


u+ 

12/  =  ma?  — 

\x-^y= 

\2x-hy-l  =  0. 


9. 


10. 


11. 


12. 


r  2/  =  3  a."^  +  wa?, 
U4-2/  +  3  =  0. 
r  2/  =  ma?  +  2Xf 
l22/-a-5=0. 

2/  =  a^  +  l, 
aa;  +  2  2/  =  10. 
a^  +  2/'  =  l, 
aa;  +  &2/  =  ^' 


2  2 

144.    Problem.   Graph  the  equation  |-  +  ^  =  1. 

Writing  the  equation  in  the  form  y  =  ±i  V25  - x^,  and  assigning 
values  to  x,  we  compute  the  corresponding  values  of  y  as  follows : 

(  x  =  0,  r  a;  =  i  5,  (  x^h  (  x  =  ~-h 


2/  =  ±4, 


\y  =  ±3.9. 


x  =  2, 
y  =  ±S.7, 


2/ =  ±3.9, 

x  =  S,  r  a:  =  4, 

2/  =  ±3.2,  t2/  =  ±2.4. 

Evidently  if  x  is  greater  than  5 
in  absolute  value,  the  correspond- 
ing values  of  y  are  imaginary. 

Plotting  these  points,  they  are 
found  to  lie  on  the  curve  shown  in 
Figure  7.  This  curve  is  called 
an  ellipse. 

EXERCISES 

Solve  the  following  pairs 
of  equations. 

In  this  way  determine  whether 
the  straight  line  and  the  curve 
intersect,  and  in  case  they  do, 

determine  the  coordinates  of  the  intersection  points.     Verify  each  by 

constructing  the  graphs. 


(0,4 

0 

ifl) 

... 

pi 

r^ 

' 

^ 

^ 

( 

-4,2 

4), 

/ 

t 

\ 

}^'' 

lA) 

i 

\ 

X-6 

0) 

(0.( 

) 

(5 

0) 

r. 

ax 

is 

/ 

( 

4,-2 

4)* 

^, 

/ 

'(4 

-2.4 

•2,-: 

^ 

^^ 

^ 

■^ 

(0,-f) 

Fig.  7. 


16+9-  ' 


!.       49  +  16-  ' 
[2x-7y  =  &. 


DISTINCT,   COINCIDENT,  AND   IMAGINARY  ROOTS     37T 


0^4- 4/  =  25,     g     f?/=2a^-3ic+4, 


2  £c  —  2/  =  4. 
3x^+2  7/=ll, 


\x- 


32/ =  7. 


25^9"^' 
x  —  y  =  5. 


y-4:X-S  =  0. 


\x  +  y  =  7. 

8.      64  +  12-'-' 
[4:y-2x=4:. 


10. 


36  +  45-^' 
-5a;+62/=10. 


49  +  25"^' 
x  +  y  =  12. 


When  arbitrary  constants  are  introduced  in  the  equations  of  a 
straight  line  and  an  ellipse,  we  may  determine  values  for 
these  constants  so  as  to  make  the  line  cut  the  ellipse,  touch  it, 
or  not  cut  it,  as  in  the  case  of  the  circle,  §  142. 

EXERCISES 

Solve  each  of  the  following  pairs  simultaneously. 

Give  such  values  to  the  constants  that  the  line  shall  (a)  cut  the 
curve  in  two  distinct  points,  (b)  be  a  tangent  to  the  curve,  (c)  have 
no  point  in  common  with  the  curve. 

In  case  (b)  is  found  very  difl&cult,  this  may  be  omitted. 


2. 


3. 


4. 


a^  +  16-   ' 

4.x-\-y  =  2. 

25  +  6^-' 
x-{-5y  =  5. 

25^16 


5. 


7. 


f^  +  ^-l 

I  16  +  25"   ' 

[ax-\-4.y=20. 

36  +  16"^' 
'aa;+6  y-24=0. 

36     25       ' 


4:X  —  5y=c.  [5  x-\- by  =  30. 

16  +  25"    '    8.       a^+4--'' 
5x-by=20.  \2x-2y=:5. 


9. 


10. 


11. 


12. 


|a^  +  2/2=r2, 

C5a^+3y'=16. 
1  hx-3y  =  S. 

fx^+7f=lU 
[x  +  by  =  12. 

far'  +  4/  =  l, 
\ax-\-  by  =  a. 


878  QUADRATIC   EQUATIONS 

SPECIAL  METHODS  OP  SOLUTION 

145.  We  have  thus  far  solved  simultaneously  one  equation 
of  the  second  degree  with  one  of  the  first  degree.  After  sub- 
stitution such  cases  reduce  to  the  solution  of  an  ordinary  quad- 
ratic, namely,  of  the  form,  aoi? -\-hx-\-c  =  0. 

While  this  is  an  effective  general  method,  yet  some  im- 
portant special  forms  of  solution  are  shown  in  the  following 
examples : 


Ex.  1.    Solve 

fa^  +  /  =  a, 

(1) 

(2) 

Square  both  members  of  (2)  and  subtract  from  (1). 

2xy  =  a-h\ 

(3) 

Add  (1)  and  (3: 
Hence 

1.     x''  +  2xy  +  y'^  =  'ia-V. 

(4) 

x  +  y  =  ii  V2a-62. 

(5) 

From  (2)  and  (5),  adding  and  subtracting 

\/2a-&2  +  j 
and 


_  V2  g  -  6^  +  & 

Xy     _  » 


V2a-&2 

^1  = ir- 


V2a-b^-h 


Ex.  2.    Solve 


I  a^  —  2/^  =  «, 
[x-y  =  b. 


(1) 

(2) 

From  (1)  (x  -y)(x  +  y)  =  a.  (3) 

Substituting  b  for  x  —  y  in  (3),   x  +  y  =  -  >  (4) 

b 

Then  (2)  and  (4)  may  be  solved  as  above. 
Ex.3.   Solve  1"  +  ^="' 


(1) 

(2) 


Multiply  (2)  by  4,  subtract  from  the  square  of  (1),  and  get 

x2-2xy-\-y^=a^~4:  b,  (3) 


whence,  x~  y  =  ±  Va^ _  4  J .  (4) 

Then  (1)  and  (4)  may  be  solved  as  in  Ex.  1, 


SPECIAL  METHODS  OF  SOLUTION  379 

x-y  =  a, 


The  equations  ,  , 

{xy=zb, 

may  be  solved  in  a  similar  manner. 

146.  We  are  now  to  study  the  solution  of  a  pair  of  equations 
each  of  the  second  degree.     See  §  6Q. 

Consider  o^  -^  y  =  a,  (1) 

x-{-f  =  b.  (2) 

Solving  (1)  for  y  and  substituting  in  (2)  we  have, 
x+a^-2ax'  +  x*  =  b, 
which  is  of  the  fourth  degree  and  cannot  be  solved  by  any 
methods  thus  far  studied.     There  are,  however,  special  cases  in 
which  two  equations  each  of  the  second  degree  can  be  solved 
by  a  proper  combination  of  methods  already  known. 

147.  Case  I.      When  only  the  squares  of  the  unknowns  enter  the 
equations. 

Example.     Solve  \    \      ,     o 

lagar  +  622/   =  Cjj. 

These  equations  are  linear  if  o^  and  y-  are  regarded  as  the 
unknowns. 

Solving  for  x^  and  y^  as  in  §  73,  we  obtain, 

Hence,  taking  square  roots, 


cM 


1"2 


-  J«1^2  -  (^9^1^  y    -_J^1^2-Q2% 

X„--  xh^2  -  (^A  fr  -   --ykA-cA 

In  this  case  there  are  four  pairs  of  numbers  which  satisfy  the  two  equa- 
tions.   This  is  in  general  true  of  two  equations  each  of  the  second  degree. 


380 


QUADRATIC  EQUATIONS 


L^^       ■^'^ 

/^ ^'^       "^^    \ 

/  /  ^       ^^^\\ 

U.l,%6)  ^  <  ^  ^         *^  \  (i>  .  2\> 

tt^/^^^'^^^twX 

hfthl       XXa%a 

f         i    J  X-axis  (0,0)        ' 

ill                       .CO 

tMTVS;    I       t/JdJ. 

Xx^t^t^^tt^tt 

V'^'NriL^  ^Jikyy--^/' 

^^^^Z-"^^'^^ 

\^  "^^           ^^ ^/ 

^^           ^^ 

Example.  Solve  simul- 
taneously, obtaining  re- 
sults to  one  decimal 
place : 


^4-^=1 
36^16 

^^-f^  25. 


a) 

(2) 


Fig.  8. 


Clear  (1)  of  fractions  and 
proceed  as  above.  Verify 
the  solution  by  reference 
to     the     graph     given     in 

Figure  8. 


EXERCISES 


Solve  simultaneously  each  of  the  following  pairs  of  equar 
tions  and  interpret  all  the  solutions  in  each  case  from  the 
graph  in  Figure  8  ; 


1.   \ 


36^16"^' 
aj2  +  2/2  =  36. 


3. 


36^10"    ' 

a^  +  /  =  49. 


^' 4.-^=1 
36      16        ' 

a^  +  f  =  16. 


4. 


^ 


r 


=1, 


36     16 

0^2+/  =  31. 


2  2 

148.   Problem.     Graph  the   equation  ^  — ^ 

2d     16 


1. 


Writing  the  equation  in  the  form  y  —  ±fVa;2  —  25,  and  assigning 
values  to  ar,  we  compute  the  corresponding  values  of  y  exactly  or 
approximately  as  follows : 

fa;=±5,  fa:  =  6i,    ja;= -6J,  j a:  =  7,  fa:=-7,      fa;=8,     ■fa:=-8, 

b=0,      b=±3,U=±3,    b=±3.9,   ly^±3.9,  ly=±5,  U=±5. 


SPECIAL  [METHODS  OF  SOLUTION 


381 


Evidently  when  x  is  less  than  5  in  absolute  value,  y  is  imaginary, 
and  as  x  increases  beyond  8  in  absolute  value,  y  continually  increases. 

Plotting  these  points,  they  are  found  to  lie  on  the  curve  as  shown  in 
Figure  9.    This  curve  is  called  a  hyperbola. 


\ 

/ 

\ 

/ 

\ 

/ 

< 

-«,6) 

\ 

t 

/ 

(8,5j 

(-7 

3.9) 

^ 

I 

/ 

W 

9) 

W 

<,5> 

\ 

/ 

^6) 

,3) 

\ 

w 

) 

(0.0 

(5.0 

z-a 

xis 

1 

) 

V 

^ 

*^y 

^A 

/ 

s 

y(6. 

*.-S) 

(-7,- 

..V 

/ 

\ 

>(7,^ 

L9) 

(■ 

«.-6). 

/ 

\ 

(8,^) 

/ 

\ 

/ 

N 

/ 

1 

\ 

Fia.  9. 


EXERCISES 

Solve  each  of  the  following  pairs  of  equations. 

Construct  a  graph  similar  to  the  one  in  Figure  8  which  shall  contain 

the  hyperbola  ■^  — ^  =  1  and  the  circles  given  in  Exs.  1,  2,  and  3. 

Construct  another  graph  containing  the  same  hyperbola  and  the 
ellipses  given  in  Exs.  4,  5,  and  6.  From  these  graphs  interpret  the 
solutions  of  each  pair  of  equations. 


1. 


16 


1, 

a^-\-y^=  16. 


2. 


25     16       ' 

.a^  +  /  =  25. 


25~16~^' 

ic2  +  /  =  36. 


382 


QUADRATIC   EQUATIONS 


4. 


25     16       ' 


+i!=i. 

36     16 


"  x^ ^  _  1 

25     16"   ' 


25     16 


25     16       ' 
16^9 


7.   Graph  the  equation  xy  =  9. 

Graph  a;?/  =  8  on  the  same  axes  with  each  of  the  following ; 
8.    a;2  4-2/2  =  16.       9.   a;2_|_2/2  ^  25.  10.   a;^  +  /  =  4. 

=  1. 


11.  -^+i!=i. 

25^16 


25     10.24 


13.    -^+^%=1. 
16^4 


14.  From  those  graphs  in  Exs.  8  to  13,  in  which  the  curves 
meet,  determine  as  accurately  as  possible  by  measurement  the 
coordinates  of  the  points  of  intersection  or  tangency. 

15.  Solve  simultaneously  the  pairs  of  equations  given  in 
Exs.  8  to  13,  after  studying  the  method  explained  in  Ex.  1, 
§  150.  Compare  the  results  with  those  obtained  from  the 
graphs. 

Note.  In  Ex.  11  clear  of  fractions  and  then  apply  §  150.  This 
gives  4:X  +  by  =  ±12  VS  and  4a:—  5y=  ±4  VS.  Solve  these 
simultaneously.  Examples  12  and  13  may  be  solved  in  a  similar 
manner. 

149.  Case  II.  Wlien  all  terms  containing  the  unkriowns  are  of 
the  second  degree  in  the  unknowns. 


Example.    Solve 


2x''-Sxy  +  4.y^  =  3, 
3x^-4:xy  +  3y^  =  2, 


Put  y  =  vxiD.  (1)  and  (2),  obtaining 

a;2(2-3?;  +  4r2)=3, 

Hence  from  (3)  and  (4), 
3 


2-3y-|-4y2' 


and  also  x^  = 


3  -  4  y  +  3  y2 


(1) 

(2) 


(3) 
(4) 


(5) 


SPECIAL   METHODS   OF   SOLUTION 

383 

Prnm  /''i^                                   '^                _                 - 

(8) 

J.rom(5)              2-3.  +  4>.2-  8_4„  +  8,^' 

or                                1)2  -  6 1)  +  5  =  0. 

(7) 

Hence                            n  =  1,  and  u  =  5. 

(8) 

From  y  =  iix,                 y  =  a:,   and  y  =  5  «. 

(9) 

li  y  =  X,  then  from  (1)  and  (2), 


^=1'   and    |^  =  -1» 


If  y  =  5ar,  then  from  (1)  and  (2), 
1 

and 


V29 


V29'  [  V2 


Verify  each  of  these  four  solutions  by  substituting  in  equations 
(1)  and  (2). 

150.  There  are  many  other  special  forms  of  simultaneous 
equations  which  can  be  solved  by  proper  combination  of  the 
methods  thus  far  used.  Also,  many  pairs  of  equations  of  a 
degree  higher  than  the  second  in  the  two  unknowns  may  be 
solved  by  means  of  quadratic  equations. 

The  suggestions  given  in  the  following  examples  illustrate 
the  devices  in  most  common  use. 

The  solution  should  in  each  case  be  completed  by  the  student. 

EX..1.   Solve  p  +  /  =  58,  0) 

Uy  =  21.  (2) 

Adding  twice  (2)  to  (1)  and  taking  square  roots,  we  have 

X  +  1/  =  10,   and  x  +  y  =  -lO.  (3) 

Each  of  the  equations  (3)  may  now  be  solved  simultaneously  with 
(2),  as  in  Ex.  3,  p.  396. 


384  QUADRATIC   EQUATIONS 


Ex.  2.  Solve 

X     y 
-1      -f 

(1) 

3  +  ^  =  13. 

(2) 

Let    -  =  a    and    -  =  6.    Then  these  equations  reduce  to 
X                    y 

a  +  5  =  5, 
■.a2  +  62  =  i3. 

(3) 
(4) 

(3)  and  (4)  may  then  be  solved  as  in  Ex.  1,  p.  378. 

Ex.3.   Solve                U  +  y'  +  x  +  y  =  i, 

(1) 

(2) 

Add  twice  (2)  to  (1),  obtaining 

x2  +  2  a:y  +  y2  +  a.  ^.  y  =  12.  (3) 

Let  X  ^-y  =  a.    Then  (3)  reduces  to 
a2  +  a  =  12, 
or,  a  =  3,  a  =  —  4.  (4) 

Hence  a?  +  y  =  3,   and  a;  +  y  =  —  4.  (5) 

Now  solve  each  equation  in  (5)  simultaneously  with  (2). 

Ex.4.   Solve'  |^y  +  a.y  =  272,  (1) 

W4-^'=10.  (2) 

In  (1)  substitute  a  for  arV-     Then 

a2  +  a  =  272,  whence  a  =  16,   and    -  17. 
Hence  xy  =  ±  Vl6=  ±  4,   and    ±  V—  17. 

Each  of  these  equations  may  now  be  solved  simultaneously  with 
(2),  as  in  Ex.  1,  p.  383. 

Ex.5.   Solve  1^-2/^  =  117,  (1) 

U-2,  =  3.  (2) 


SPECIAL  METHODS  OE  SOLUTION  885 

By  factoring,  (1)  becomes 

(x-yXx^-\-xy-{.y^)=n7.  (3) 

Substituting  3  for  a:  —  y,  we  have 

x^  +  xy-\-y^  =  39.  (4) 

(2)  and  (4)  may  now  be  solved  by  substitution  as  in  §§  140-144. 

Ex.6.   Solve  1-^  +  ^  =  513,  (1) 

U  +  .V  =  9.  (2) 

Factor  (1)  and  substitute  9  for  a:  +  y.    Then  proceed  as  in  Ex.  5. 

Ex.7.   Solve  (^  +  =^  =  126,  (1) 

U  +  y=9.  (2) 

Factoring  (1)  and  substituting  9  for  a:  +  y,  we  have 

xy  =  U.  (3) 
(2)  and  (3)  may  then  be  solved  as  in  Ex.  3,  p.  378. 


Ex.8.   Solve  fx»  +  /  =  54x2,, 


(1) 

(2) 


Factor  (1)  and  substitute  6  for  x-\-  y,  obtaining 

x^-  xy-\-  y^  =  9  xy.  (3) 

(2)  and  (3)  may  now  be  solved  by  substitution,  as  in  §§  140-144. 


I 


Ex.9.    Solve               ^^-f  =  eS,  (1) 

x'  +  xy  +  f  =  21,  (2) 

Factor  (1)  and  substitute  21  for  x^  -\-  xy  +  y%  then  proceed  as  in 

Ex.8. 

Ex.10.    Solve             {^  +  2/^  =  243,  (1) 

\a^y-{-xy'  =  162.  (2) 

Multiply  (2)  by  3  and   add  to   (1),  obtaining  a    perfect  cube. 
Taking  cube  roots,  we  have 

x-hy  =  9.  (3) 

(1)  and  (3)  are  now  solved  as  in  the  preceding  example. 


386 


QUADRATIC  EQUATIONS 


Ex.  11.    Solve  |-^  +  2/*  =  641.  W 

[x  +  y  =  7.  (2) 

Raise  (2)  to  the  fourth  power  and  subtract  (1),  obtainiug 

4  x^y  +  QxY  +  4  xy^  =  1760.  (3) 

Factoring,       2  xy  (2  x^ -\- 3  xy -{- 2  y^)  =  1760.  (4) 

Squaring  (2)  we  have 

2x2 +  4x2/ +  2  2/2  =  98,  (^) 

or  2x^  +  dxy  +  2y^  =  9S-xy,  (6) 

Substituting  (6)  in  (4),  we  have 

2  xy  (98  -  xy)  =  1760,  •         (7) 

or  a;22/2- 98x2/ +  880  =  0.  (8) 

In  (8)  put  xy  =  a,  obtaining 

a2_  98a +  880  =  0.  (9) 

The  solution  of  (9)  gives  two  values  for  xy,  each  of  which  may  now 
be  combined  with  (2)  as  in  Ex.  3,  p.  378. 

EXERCISES 

Solve  each  of  the  following  pairs  of  equations : 


1. 


2. 


3. 


4. 


13. 


14. 


{'- 


3. 


5. 


9.     i 


3a^ +-2  2/2  =  35, 
2a;2-32/'  =  6. 

3x'  +  2xy  =  16, 
4x2-3a72/  =  10. 

a2+-a6+-62  =  7, 
a'-ab  +  b'  =  m 

(Sx-2y  =  6, 
\Sx^-2xy  +  4.f  =  12. 

(a-}-b-]-ah  =  n, 
[(a  +  by  +  a'b'  =  61. 


[xy  =  2. 

ra^-+2/2  =  5,     10.    {-'^  +  2/^  =  11 


U'3  +  2/3  =  91, 
[x-\-y  =  7. 


7. 


8. 


a;2  +  z2  =  10, 

2/2  +  2;2=13.    11. 

Sx-4:y=0, 


(Sx- 


f 


4. 


x'^-\-xy  =  4, 


(^  4"  -^^  —  ^j 
y^  +  xy=5. 


12. 


I  0^2-2/^  =  7. 
(x^-3xy=0, 

i  +  i  =  19, 

1+1=1. 

X     y 


15. 


1+1+ 1=49, 


17. 


18. 


19. 


20. 


22. 


23. 


24. 


25. 


26. 


37. 


SPECIAL   METHODS   OF   SOLUTION  387 

^a'-2ab  =  b'-16,  ^^      f  (a;- 4)^  + (2/ +  4)2  =100, 


(4:a'-2ab  =  b'- 
'     [5d'  =  7ab-S6. 


3a^-9/=12, 
2x-3y  =  U. 

{x^  +  f  =  3. 


[xy  = 


a^^f  +  x  +  y  =  lS, 
6. 


'x^-^f-\-x-y  =  S6, 
xy  =  15. 


(x^-5xy  +  f=-2, 

•     \x^-^7xy  +  f-  =  22. 


(a^-\-6ab-^b'  =  124:, 
[a  +  b  =  S. 

(a'-Sab-{-2b'  =  0, 
|2a2  +  a6-6*  =  9. 

'x'  +  f-  +  2x  +  2y  =  27, 
xy  =  -12. 


27. 


28. 


29. 


30. 


31. 


32. 


33. 


34. 


or  -^  y-  —  5  X  —5  y  =  —  4:,   35 
xy  =  5. 


r(7  +  rr)(64- 
I  a;  +  2/  =  5. 


[ic+y  =  14. 

ra;2/  +  2/  +  a;  =  17, 
1  ar'y^  ^y2_^^^  129. 

6  +  a2  =  5(a-^), 
a  +  62^2(a-6). 


+  22/2  =  177, 
13a;2^3^ 


r(i3a^)2H 

1(2  2/)^- 


9^25 


Cf)' 


16, 


fx2  +  2/2  =  20, 
l5ar^-32/'^  = 


7/)  =  80, 


36.     ^ 


28. 


'  0^  =  —  5  —  3  a^, 
^2aJ2/  =  /-24. 

•«  +  2/  +  Va;  +  2/  =  12, 
ar'  +  2r^  =  189. 

x'-^x'y'  +  y^^lSS, 
xF  —  xy  +  y^  =  7. 

'x  +  xy-\-y  =  29, 
^ar-{-xy  +  y'^  =  61. 


(2x'-5xy  +  3x-2y  =  22, 
\5xy  +  7x-Sy-2x^  =  S. 


39. 


ra;  +  2/  =  74, 
la^  +  2/='  =  3( 

=  29, 

7  2/2-17  2/= -17. 


3026. 


r  72/2-5 ar^  + 20a.' +  132/  =  29, 
|5(a;-2)2- 


388  *       QUADRATIC  EQUATIONS 

(Sx  +  4.y)(7x-2y)-7x  +  2y  =  30, 


40.     I 


[  x^ -\- x^y  -\- XI 


x'y  +  xy'  +  f  =  32. 

^-f  =  31,  ^^      {x'  +  f-xy  =  %Q, 

44.     \ 
x  —  y  =  l,  ix  —  y  —  xy  =  —  S. 


a^  +  x'y-^xf  +  y'  =  120, 


5b-b^-ab  =  24:. 


»■{ 


x^  —  afy-{-  xy^  —  y^  =  40. 


2(x  +  4.y-5{y-7y=75, 
'    "*  7  (x  + 4)2 +15  (2/ -7)2  =1075. 

la^  —  xy  +  y^  =  a—b. 


HIGHER  EQUATIONS  INVOLVING  QUADRATICS 

151.  An  equation  of  a  degree  above  the  second  may  often  be 
reduced  to  the  solution  of  a  quadratic  after  applying  the  factor 
theorem.     See  §  92. 

Example.     Solve       2a^-{-ic^ -10x-\-7  =  0.  (1) 

By  the  factor  theorem,  a;  —  1  is  found  to  be  a  factor, 

giving  (x-l)(2x^-\-dx-7)=0.  (2) 

Hence  by  §  22,  a:  -  1  =  0  and  2  a:2  +  3  x  -  7  =  0.  (3) 

From  X  -  1  =  0,  x  =  1.  (4) 

From2a:2  +  3a;-7  =  0,         x  =  ~  ^  ^^^ .  (5) 

Hence  (4)  and  (5)  give  the  three  roots  of  (1). 


HIGHER   EQUATIONS  INVOLVING   QUADRATICS       389 
EXERCISES 

Solve  each  of  tlie  following  equations : 

1.  7a^-lla^+4:X^0.  5.   2Sa:^-10x'-Ux  =  6* 

2.  3x*+a^-}-2ay^  +  24:X  =  0.          6.   a^-3x^  +  Sa^-x  =  0. 

3.  3a53-16ar»4-23aj-6=0.         7.   4:a^  +  12x^-Sx-9=0. 
A.   5x^+2x^  +  4:X=-7..  8.   x*-5a^-{-2a^+20x=24:, 

9.   6x^-{-29a^-19x  =  16. 
10.   15a;*  +  49a^-.92a^  +  28x  =  0. 

EQUATIONS  IN  THE  FORM  OF  QUADRATICS 

152.  If  an  equation  of  higher  degree  contains  a  certain 
expression  and  also  the  square  of  this  expression,  and  involves 
the  unknown  in  no  other  way,  then  the  equation  is  a  quadratic 
in  the  given  expression. 

Ex.1.    Solve  iB*  +  7a*  =  44.  (1) 

This  may  be  written,  (x^y  -\-  7(x^)  =  44,  (2) 

which  is  a  quadratic  in  x^.     Solving,  we  find 

a;2  =  4  and  x^  =  -  11.  (3) 

Hence,  x  =  ±2  and  x  =  ±  V-11.  (4) 


Ex.2.    Solve  ic  +  2  +  3V^+2  =  18.  (1) 


Since  a:  +  2  is  the  square  of  Vx  -\-  2,  this  is  a  quadratic  in  Vx  +2. 
Solving  we  find        Va:  +  2  =  3  and  Vx  +  2  =  -  6.  (2) 

Hence  a;  +  2=  9  and  a:+2  =  36,  (3) 

Whence  x  =  7  and  a:  =  34.  (4) 

Ex.  3,    Solve   (2a^-l)2_5(2a;2-l)-14  =  0. 

First  solve  as  a  quadratic  iu  2  x^  ~  1  and  then  solve  the  two  result- 
ing quadratics  in  x. 


390  QUADRATIC   EQUATIONS 


Ex.4.   Solve   a^-7x  +  A0-2^/x'-7x  +  6^=-26,        (1) 
Add  29  to  each  member,  obtaining 

a;2  _  7  X  +  69  -  2Vx2-7a;  +  69=  3.  (2) 


Solve  (2)  as  a  quadratic  in  Vx^  —  7x  +  69,  obtaining 


V2:2  -  7a:  +  69  =  3  and  Vx2  -  7  x  +  69  =  -  1,  (3) 

whence  a;^  -  7  a:  +  69  =  9  or  1.  (4) 

The  solution  of  the  two  quadratics  in  (4)  will  give  the  four  values 
of  X  satisfying  (1). 

EXERCISES 

Solve  the  following  equations  : 


1.   af  +  2a^  =  S0.  2.   5a;-4-2V5a;-4  =  63. 

3.    (2-x  +  a^'-\-x^-x  =  lS. 


4.  a2-3a  +  4-3Va2-3a  +  4=-2. 

5.  3a«-7a3-1998  =  0. 


6.   a^-8aj  +  16  +  6Va^-8a;  +  16  =  40. 
8.   a«  -  97  a^  + 1296  =  0. 


9.   a2-3a  +  4  +  Va2-3a  +  15  =  19. 
10.    (5x-T-j-Sxy  +  3x'-\-5x-2^7  =  0. 
n.    ■^7aj-6-4^/7a;-6  +  4  =  0. 


RELATIONS    BETWEEN  THE  ROOTS  AND  THE  COEFFICIENTS  OR  A 

QUADRATIC 

153.   If  in  the  general  quadratic,  ax^  +  bx-{-c  =  0,  we  divide 

b  c 

both  members  by  a  and  put  -=p,   -  =  g,  we  have  x^-\-px-{-q=0. 

a  a 


Solving,  x.=  -P  +  y-^?,  and  ^=-P-^/-'^1 


RELATIONS  OF  COEFFICIENTS   AND   ROOTS  391 

Adding  ajj  and  x^,  jCi  +  iCg  = ^=  —p,  (1) 

Multiplying  Xi  and  Xz,       x^x^  =  ^  ~\^~ — it  =  g.  (2) 

Hence  in  a  quadratic  of  the  form  x^-\-px-\-q  =  0,  the  sum  of 
the  roots  is  —  p,  and  the  product  of  the  roots  is  q. 

The  expression p'^-4.q=^-^  =  b'-^ac 
^  ^  ^     a^      a  a' 

Hence  p^  —  Aq  is  positive,  negative,  or   zero,   according  as 

l^  —  4:ac  is  positive,  negative,  or  zero. 

Hence,  as  found  on  pp.  387,  389,  the  roots  of 

ax^  +  bx  -\-  c  =  0,  or  Qc^  -{-  px  -{-  q  =  0  are : 

real  and  distinct,  ifb^  —  4:ac>0,      or  p^  —  4:q>0,  (3) 

real  and  equal,  if    6^  —  4  ac  =  0,      or  p^  —  4  g  =  0,  (4) 

imaginai'y,  if  6^  —  4  ac  <  0,      or  p^  —  4:q  <  0.  (5) 

By  means  of  (1)  to  (5),  we  may  determine  the  character  of 
the  roots  of  a  quadratic  without  solving  it. 

Ex.  1.   Determine  the  character  of  the  roots  of 
8a.'2-3a;-9  =  0. 

Since  62_4ac=9-4  •  8(-9)=297>0,  the  roots  are  real  and  dis- 
tinct.    Since  b^  —  i  ac  is  not  a.perfect  square,  the  roots  are  irrational. 

Since  ^  =  —  f  =  x^x^,  the  roots  have  opposite  signs. 

Since  p  =  —  |  or  —  jt?  =  |  =  Xj  +  Xg,  the  positive  root  is  greater  in 
absolute  value. 

Ex.  2.  Examine  3a^  +  5aj  +  2=0. 

Since  J^  _  4  ac  =  25  —  4  •  3  •  2  =  1  >  0,  the  roots  are  real  and  distinct. 

Since  b^  —  4:  ac  is  a  perfect  square,  the  roots  are  rational. 

Since  5  =  |  =  XiX2,  the  roots  have  the  same  sign. 

Since  —  p  =  —  |  =  a:^  +  Xg,  the  roots  are  both  negative. 

Ex.  3.   Examine  ar^  —  14  a;  +  49  ^c  0. 

Since  jo^  —  4  5  =  196  —  4  •  49  =  0,  the  roots  are  real  and  coincident 

Ex.  4.   Examine  a;^  —  7  a;  +  15  =  0. 

Since  ^j^  —  4^  =  49  —  4.15=  —ll,  the  roots  are  imaginary. 


392  QUADRATIC   EQUATIONS 

EXERCISES 

Without  solving,  determine  the  character  of  the  roots  in  each 
of  the  following : 

1.  5a^-4aj-5  =  0.  9.  16  m^ -|- 4  =  16  m. 

2.  6a^  +  4i»  +  2  =  0.  10.  25a'-10a  =  S. 

3.  a;2-4a;  +  8  =  0.  U.  20  -  13b -15b^  =  0. 

4.  2  +  2aj2  =  4a;.  12.  lOy^ -{- S9y +  U  =  0, 

5.  6a;  +  8a;2  =  9.  13.  Sa^^.  5a +  22  =  0. 

6.  l-a^  =  Sa.  14.  3a^  -  22a  +  21  =  0. 

7.  6a-30  =  3a2.  15^  5  6^ -f  6  6  =  27. 

8.  6 a^  4- 6  =  13 a.  16,  6a  -  17==  11  a^ 

FORMATION  OP  EQUATIONS  WHOSE  ROOTS  ARE  GIVEN 

154.  Ex.  1.   Form  the  equation  whose  roots  are  7  and  —  4. 

From  (1)  and  (2),  §  153,  we  have 

a:j  +  ^2  =  -  p  =  7  +  (-  4)  =  3.     Hence/)  =  -  3. 
And  arjXg  =  ^  =  7(-  4)  =  -  28, 

Hence         x^  -{■  px  -\-  q  =  0  becomes  a:^  —  3  ar  —  28  =  0. 

In  case  the  equation  is  to  have  more  than  two  roots,  we  pro- 
ceed as  in  the  following  example : 

Ex.  2.  Form  the  equation  whose  roots  are  2,  3,  and  5, 

Recalling  the  solution  by  factoring,  we  may  write  the  desired  equa- 
tion in  the  factored  form  as  follows : 

(a;-2)(a:-3)(a:-  5)  =0. 

Obviously  2,  3,  and  5,  are  the  roots  and  the  only  roots  of  this  equa- 
tion.    Hence  the  desired  equation  is : 

(a;  -  2)(a;  -  3)(a:  -  5)  =  a;8  -  10a:2  +  31  a:  -  30  =  0. 


EQUATIONS   WITH   GIVEN  ROOTS  393 
EXERCISES 

Form  the  equations  whose  roots  are : 

1.  3,-7.                   4.   5,  -  4,  -  2.  7.    -5,  -6. 

2.  h,c.                         6.    V6,  -V6.  8.    -h-\-k,  -h-k. 

3.  a,  -  6,  -  c.       6.   a  —  V3,  a  +  V3.  9.  V^^,  ~  V^^. 
10.  a,  -  h.            11.   8  +  V3,  8  -  V3.  12.  2,  3,  4,  5. 
13.  3  +  2  V^^,  3-2  V^=3.           14.  5  -  V^^,  5  +  V^^. 


15.    l,4,i,3.  16. 


5^-V6=^-4ac    -5-V62-4ac 


^'^'"*  *"•  2a  '  2a  * 

155.  An  expression  of  the  second  degree  in  a  single  letter 
may  be  resolved  into  factors,  each  of  the  first  degree  in  that 
letter,  by  solving  a  quadratic  equation. 

Ex.  1.   Factor  6ar^  -  17  a;  +  5. 

This  trinomial  ihay  be  written,  6(3:^  —  J^  ar  +  |). 

Solving  the  equation,  x^  —  Y  ^^  +  ^  =  0,  we  find  x^  —  \  and  x^  =  f . 
Hence  by  the  factor  theorem,  §  92,  a;  —  ^  and  a:  —  f  are  factors  of 
a;2  _  Y-x  +  |.     And  finally 

6(xa  -  17  a:  +  5)  =  6(x  -  I) (a:  -  f )  =  3 (X  -  i)  .  2(a;  -  O 
=  (3  2:-l)(2x  — 5). 

This  process  is  not  needed  when  the  factors  are  rational,  but 
it  is  applicable  equally  well  when  the  factors  are  irrational  or 
imaginary. 

Ex.2.  Factor  3a^  +  8a;-7  =  3(a^  +  |a;-|). 

Solving  the  equation  x^  +  |  a:  —  ^  =  0,  we  find, 

^3  ^3 

Hence  as  above : 

3x'  +  8x-7  =  3[x--t+^][.--^;^] 


394  QUADRATIC   EQUATIONS 

EXERCISES 

In  exercises  1  to  16,  p.  410,  transpose  all  terms  of  each  equar 
tion  to  the  first  member,  and  then  factor  this  member. 

PROBLEMS   INVOLVING    QUADRATIC   EQUATIONS 

In  each  of  the  following  problems,  interpret  both  solutions 
of  the  quadratic  involved  : 

1.  The  area  of  a  rectangle  is  2400  square  feet  and  its  perim- 
eter is  200  feet.     Pind  the  length  of  its  sides. 

2.  The  area  of  a  rectangle  is  a  square  feet  and  its  perimeter 
is  2  6  feet.  Find  the  length  of  its  sides.  Solve  1  by  substitu- 
tion in  the  formula  thus  obtained. 

3.  A  picture  measured  inside  the  frame  is  18  by  24  inches. 
The  area  of  the  frame  is  288  square  inches.     Find  its  width. 

4.  If  in  problem  3  the  sides  of  the  picture  are  a  and  b  and 
the  area  of  the  frame  c,  find  the  width  of  the  frame. 

5.  The  sides  a  and  &  of  a  right  triangle  are  increased  by  the 
same  amount,  thereby  increasing  the  square  on  the  hypotenuse 
by  2  k.     Find  by  how  much  each  side  is  increased. 

Make  a  problem  which  is  a  special  case  of  this  and  solve  it  by  sub- 
stitution in  the  formula  just  obtained. 

6.  The  hypotenuse  c  and  one  side  a  are  each  increased  by 
the  same  amount,  thereby  increasing  the  square  on  the  other 
side  by  2  k.    Find  how  much  was  added  to  the  hypotenuse. 

Make  a  problem  which  is  a  special  case  of  this  and  solve  it  by 
substituting  in  the  formula  just  obtained. 

7.  A  rectangular  park  is  80  by  120  rods.  Two  driveways  of 
equal  width,  one  parallel  to  the  longer  and  one  to  the  shorter 
side,  run  through  the  park.  What  is  the  width  of  the  drive- 
ways if  their  combined  area  is  591  square  rods  ? 

8.  If  in  problem  7  the  park  is  a  rods  wide  and  b  rods  long 
and  the  area  of  the  driveways  is  c  square  rods,  find  their 
width. 


PROBLEMS  INVOLVING   QUADRATICS  395 

9.   The  diagonal  of  a  rectangle  is  a  and  its  perimeter  2  b. 
Find  its  sides. 

Make  a  problem  which  is  a  special  case  of  this  and  solve  it  by  sub- 
stituting in  the  formula  just  obtained. 

10.  If  in  problem  9  the  difference  between  the  length  and 
width  is  b  and  the  diagonal  is  a,  find  the  sides.  Show  how 
one  solution  can  be  made  to  give  the  results  for  both  problems 
9  and  10. 

11.  Find  two  consecutive  integers  whose  product  is  a. 

Make  a  problem  which  is  a  special  case  of  this  and  solve  it  by  sub- 
stituting in  the  formula  just  obtained. 

What  special  property  must  a  have  in  order  that  this  problem  may 
be  possible.     Answer  this  from  the  formula. 

12.  A  rectangular  sheet  of  tin,  12  by  16  inches,  is  made  into 
an  open  box  by  cutting  out  a  square  from  each  corner  and 
turning  up  the  sides.  Find  the  size  of  the  square  cut  out  if 
the  volume  of  the  box  is  180  cubic  inches.    • 

The  resulting  equation  is  of  the  third  degree.  Solve  it  by  factor- 
ing. See  §  151.  Obtain  three  results  and  determine  which  are  appli- 
cable to  the  problem. 

13.  A  square  piece  of  tin  is  made  into  an  open  box  contain- 
ing a  cubic  inches,  by  cutting  from  each  corner  a  square  whose 
side  is  b  inches  and  then  turning  up  the  sides.  Find  the 
dimensions  of  the  original  piece  of  tin. 

14.  A  rectangular  piece  of  tin  is  a  inches  longer  than  it  is  wide. 
By  cutting  from  each  corner  a  square  whose  side  is  b  inches  and 
turning  up  the  sides,  an  open  box  containing  c  cubic  inches  is 
formed.     Find  the  dimensions  of  the  original  piece  of  tin. 

15.  The  hypotenuse  of  a  right  triangle  is  20  inches  longer 
than  one  side  and  10  inches  longer  than  the  other.  Find  the 
dimensions  of  the  triangle. 

16.  If  in  problem  15  the  hypotenuse  is  a  inches  longer  than 
one  side  and  b  inches  longer  than  the^other,  find  the  dimen- 
sions of  the  triangle. 


396  QUADRATIC   EQUATIONS 

17.  The  area  of  a  circle  exceeds  that  of  a  square  by  10 
square  inches,  while  the  perimeter  of  the  circle  is  4  less  than 
that  of  the  square.  Find  the  side  of  the  square  and  the  radius 
of  the  circle. 

Use  3}  as  the  value  of  tt. 

18.  If  in  problem  17  the  area  of  the  circle  exceeds  that  of 
the  square  by  a  square  inches,  while  its  perimeter  is  2  &  inches 
less  than  that  of  the  square,  find  the  dimensions  of  the  square 
and  the  circle. 

Determine  from  this  general  solution  und6r  what  conditions  the 
problem  is  possible. 

19.  Find  three  consecutive  integers  such  that  the  sum  of 
their  squares  is  a. 

Make  a  problem  which  is  a  special  case  of  this  and  solve  it  by 
means  of  the  formula  just  obtained.  From  the  formula  discuss  the 
cases,  a  =  2,  a  =  5,  a  =  14.  Find  another  value  of  a  for  which  the 
problem  is  possible. 

20.  The  difference  of  the  cubes  of  two  consecutive  integers 
is  397.     Find  the  integers. 

21.  The  upper  base  of  a  trapezoid  is  8  and  the  lower  base  is 
3  times  the  altitude.  Find  the  altitude  and  the  lower  base  if 
the  area  is  78. 

See  problem  7,  p.  326. 

22.  The  lower  base  of  a  trapezoid  is  4  greater  than  twice 
the  altitude,  and  the  upper  base  is  ^  the  lower  base.  Find 
the  two  bases  and  the  altitude  if  the  area  is  62^. 

23.  The  lower  base  of  a  trapezoid  is  twice  the  upper,  and  its 
area  is  72.  If  i  the  altitude  is  added  to  the  upper  base,  and 
the  lower  is  increased  by  J  of  itself,  the  area  is  then  120. 
Find  the  dimensions  of  the  trapezoid. 

24.  The  upper  base  of  a  trapezoid  is  equal  to  the  altitude, 
and  the  area  is  48.  If  the  altitude  is  decreased  by  4,  and  the 
upper  base  by  2,  the  ar^a  is  then  14.  Find  the  dimensions  of 
the  trapezoid. 


PROBLEMS  INVOLVING   QUADRATICS  397 

25.  The  upper  base  of  a  trapezoid  is  4  more  tlian  ^  the 
lower  base,  and  the  area  is  84.  If  the  upper  base  is  decreased 
by  5,  and  the  lower  is  increased  by  |  the  altitude,  the  area  is 
78.     Find  the  dimensions  of  the  trapezoid. 

26.  The  area  of  an  equilateral  triangle  multiplied  by  V3, 
plus  3  times  its  perimeter,  equals  81.  Find  the  side  of  the 
triangle. 

See  problem  15,  p.  191,  E.  C. 

27.  The  area  of  a  regular  hexagon  multiplied  by  V3,  minus 
twice  its  perimeter,  is  504.     Find  the  length  of  its  side. 

See  problem  20,  p.  192,  E.  C. 

28.  If  a  times  the  perimeter  of  a  regular  hexagon,  plus  V3 
times  its  area,  equals  b,  find  its  side. 

29.  The  perimeter  of  a  circle  divided  by  tt,  plus  V3  times 
the  area  of  the  inscribed  regular  hexagon,  equals  122^.  Find 
the  radius  of  the  circle. 

80.  The  area  of  a  regular  hexagon  inscribed  in  a  circle  plus- 
the  perimeter  of  the  circle  is  a.     Find  the  radius  of  the  circle. 

31.  One  edge  of  a  lectangular  box  is  increased  6  inches, 
another  3  inches,  and  the  third  is  decreased  4  inches,  making 
a  cube  whose  volume  is  864  cubic  inches  greater  than  that  of 
the  original  box.     Find  its  dimensions. 

32.  Of  two  trains  one  runs  12  miles  per  hour  faster  than  the 
other,  and  covers  144  miles  in  one  hour  less  time.  Find  the 
speed  of  each  train. 

In  a  township  the  main  roads  tun  along  the  section  lines,  one  half 
of  the  road  on  each  side  of  the  line. 

33.  Find  the  area  included  in  the  main  roads  of  a  township 
if  they  are  4  rods  wide. 

34.  If  the  area  included  in  the  main  roads  of  a  township  is 
68,796  square  rods,  find  the  width  of  the  roads. 

35.  Find  the  width  of  the  roads  in  problem  34  if  the  area 
included  in  them  is  a  square  rods. 


CHAPTER  VIII 

ALGEBRAIC  FRACTIONS 

156.   An  algebraic  fraction  is  the  indicated  quotient  of  two 

algebraic  expressions. 

n 
Thus  -  means  n  divided  by  d. 

From  the  definition  of  a  fraction  and  §  11,  it  follows  that 
{hQ  product  of  a  fraction  and  its  denominator  equals  its  numerator. 

That  is,  ^  d-^=^n. 

a 

REDUCTION  OF  FRACTIONS 

.  157.  The  form  of  a  fraction  may  be  modified  in  various  ways 
without  changing  its  value.  Any  such  transformation  is  called 
a  reduction  of  the  fraction. 

The  most  important  reductions  are  the  following : 

(A)   By  manipulation  of  signs. 


E.g. 

n_       —n_        n    _— w.    b  —  a_      a  —  h      a  —  h 
d            d             —  d      —  d'    c  —d          c  -  d     d  —  c 

(-B) 

To  lowest  terms. 

E.ff. 

a;4  +  a;2+l                    (x^  +  a:  +  1)('j:2  -  a:  +  1) 

1 

(a:-l)(x  +  l) 
(C)  To  integral  or  mixed  expressions. 

^  x2  +  1  a;2  +  1  a;2  +  1 


REDUCTION  OF  FRACTIONS  399 

(D)  To  equivalent  fractions  having  a  common  denominator. 

E.q.  — —   and   become  respectively  ^^  "^    ' and 

— ^^^  "^  ^) — ;  a  +  1  and  — i-  become  respectively  ^^  ~\  and  — i-  • 

158.  These  reductions  are  useful  in  connection  with  the 
various  operations  upon  fractions.  They  depend  upon  the 
principles  indicated  below. 

Reduction  (^)  is  simply  an  application  of  the  law  of  signs  in 

division,  §  28.    It  is  often  needed  in  connection  with  reduction  (D). 

See  §  159. 

Reduction  (JB)  depends  upon  the  theorem,  §  47,  77  =  7,  by  which 

ok     0 

a  common  factor  may  he  removed  from  both  terms  of  a  fraction.     It  is 

useful  in  keeping  expressions  simplified.     This  reduction  is  complete 

when  numerator  and  denominator  have  been  divided  by  their  H.  C.  F. 

See  §§  95-102. 

Reduction  (C)  is  merely  the  process  of  performing  the  indicated 
division,  the  result  being  integral  when  the  division  is  exact,  otherwise 
a  mixed  expression. 

In  case  there  is  a  remainder  after  the  division  has  been  carried 
as  far  as  possible,  this  part  of  the  quotient  can  only  be  indicated. 

Thus        .  f  =  9  +  f'. 

in  which  D  is  dividend,   d  is  divisor,    q  is   quotient,   and  i?  is 
remainder. 

Reduction  (Z))  depends  upon  the  theorem  of  §  47,  7  =  t? >  by  which 

a  common  factor  is  introduced  into  the  terms  of  a  fraction. 

A  fraction  is  thus  reduced  to  another  fraction  whose  denominator 
is  any  required  multiple  of  the  given  denominator. 

If  two  or  more  fractions  are  to  be  reduced  to  equivalent  fractions 
having  a  common  denominator,  this  denominator  must  be  a  common 
multiple  of  the  given  denominators,  and  for  simplicity  the  L.  C.  M.  is 
used. 


400  ALGEBRAIC  FRACTIONS 


EXERCISES 

Reduce  the  following  so  that  the  letters  in  each  factor  shall 
occur  in  alphabetical  order,  and  no  negative  sign  shall  stand 
before  a  numerator  or  denominator,  or  before  the  first  term  of 
any  factor. 

1    ^  — ^,  ly    —(c'-a)(d  —  c)^ 

b-a*  '     (a-6)(6-c) 

2.       (b-a)(c-d)  g    (b-a)(c-b)(c-a) 

3      -(g^-y)    .  9^ ri 

(b  -  a)(c-d)  (a  -  b)(b  -c){c-  a) 

4    -(^-y)(^-y)  10.    (c--b-a)(b-a-c)_ 

-(6-a)(c-d)  3(a-c)(6-c)(c-a) 


r  — 


11. 


(Sc^2a)(4.b-a)d 

(a  —  b)(c  —  b){c  —  a)  '  (-a  +  6)(a-6)(c-a)* 

b(c~a)  '     (w  — m)(— A;  — m  — Q 

Reduce  each  of  the  following  to  lowest  terms : 

13  «'-^'  18      ^  +  2x'-\-2x-^l 

14  (^-(a-bY  j9     2a^-ar^-8a;-3  ^ 
(a  +  c)2-62'  '  2»3_3ic2_7a;  +  3* 

7aar'-56aV  4  a^  4-83^^-30;  4-5 

'  28x2(l-64aV)*  '  6iC»-5a5«  +  4a)-.l' 

16  ^^4-5m^  +  7m4-3  ^^      a^  — a^4-y'4-g?— y-f3 

^2  +  4m4-3       *  *  a^-f  ^4-a'-2-r  +  3ajH-32/' 

17  a^-Ta  +  6  ^^    a^ -\- a'b'' -{- b' -h a^ -\- b\ 
^'  a^^7a^  +  Ua-S'  '     a' -\- ab -\- b' -{- a  +  b 


REDUCTION  OF  FRACTIONS  401 

23. 


X'* -\- 4:  a^y -\- 6  a^y^  +  4  xy'^  +  y*  —  a* 
a^-\-2xy  +  f  —  a^ 


84. 


25. 


36. 


27. 


3?y  —  7?z  +  yh  —  xy^-\-  xz^  —  ys^ 

2x*-x'-20ay^-^16x-S 
3  a.**  +  6  a^  ~  30  ic«  -  41  a;  + 15* 


Eeduce  each    of   the   following  to  an  integral  or  mixed 
expression : 


28.  ^^^-^• 
a;  +  l 

30. 

29.   ^-^}. 

81. 

,,    a'  +  a'b'-{-b* 

Ol.                                         -• 

a  —  b. 

^^    3a»-3a'  +  3a 

-1 

a«  +  a  +  l 
36. 


•  c»  +  c2-c  +  l 

33.  ^-"  +  .^- 
x'  +  x-^l 

a!«- 

_iB2_a;  +  l 

i»*  +  cc«  +  a;  — 1 

47W 

,4_3^S4.3 

a-2  ^^'   2m'-2m4-l 

Reduce  each  of  the  following  sets  of  expressions  to  equivar , 
lent  fractions  having  the  lowest  common  denominator ; 


38. 


39. 


a;*  — 3ar^/  +  2/*'    x^  —  xy  —  y^'    y?-\-xfy  —  jf 

a4-6 CT  b 

5a2c  +  12cd-6a(i-10ac2'    ^ac-^d'    a-2c' 


40  a?^  +  y^  a;  +  y'-l       «!±^±y!. 

•  a^  +  f  +  x'-xy  +  f   x'-^xy^f'     x^y  +  1  ' 


402  ALGEBRAIC   FRACTIONS 

^j    X y 

*  (a-b)(c-b)(c-ay    (a- b)(b-c)(a- c)' 

42.  -^,  -±-,  -^,  d.      r - •• 

43  &  — c    '  a  —  b  c  —  a 

*  (a_c)(a-6)'    (c-a)(6-c)'    (6_a)(c-6)' 

^ .    m  —  n a-\-2  ct  +  3 

*  a^-Ga^  +  lla-e'    a2-4a  +  3'    o?-^a  +  2 

If  a,  6,  m  are  positive  numbers,  arrange  each  of  the  follow- 
ing sets  in  decreasing  order.  Verify  the  results  by  substitut- 
ing convenient  Arabic  numbers  for  a,  b,  m. 

Suggestion.  Reduce  the  fractions  in  each  set  to  equivalent  frac- 
tions having  a  common  denominator. 

._        a          2a         Sa  ..         m  2m  3m 

45.  -,    -,    -•       46. 


a  +  1'    a  +  2'    a  +  3  *  2m  +  l'    3m  +  2'    4m  +  3 

^„    a-\-Sb      a-\-b      a  +  46 
*  a-}-46'    a-\-2b'    a-{-5b' 

48.    Show  that,  for  a  different  from  zero,  neither  ^"^^  nor 

n—a  n  ^+" 

can  equal  -,  unless  n  =  d.     State  this  result  in  words, 

d  —  a  d 

and  fix  it  in  mind  as  an  impossible  reduction  of  a  fraction. 


ADDITION  AND  SUBTRACTION  OF  FRACTIONS 

159.  Fractions  which  have  a  common  denominator  are  added 
or  subtracted  in  accordance  with  the  distributive  law  for 
division,  §§  30,  31. 

That  is,  g     6_c^a  +  &-c. 

'  d     d     d  d 

In  order  to  add  or  subtract  fractions  not  having  a  common 
denominator,  they  should  first  be  reduced  to  equivalent  frac- 
tions having  a  common  denominator. 


ADDITION  AND   SUBTRACTION   OF   FRACTIONS       403 

When  several  fractions  are  to  be  combined,  it  is  sometimes 
best  to  take  only  part  of  them  at  a  time.  In  any  case  it  is 
advantageous  to  keep  all  expressions  in  the  factored  form  as 
long  as  possible. 

1  1  .  1 


Ex. 


(x  -  l){x  -  2)      (2  -  x)(x  -  3)      (3  -  x)(4  -  x) 


Taking  the  first  two  together,  we  have 
1  .  1  2x-4: 


(a:-l)(x-2)      (x-2)(x-S)      (a;-l)(a:-2)(a;-3)       (x-l)(a,-3) 

Taking  this  result  with  the  third, 

2  .  1  3a:-9  3 


(x-l)(x-S)      (a;-3)(a:-4)      (a:-l)(a:-3)(a:-4)       (x-l){x-^) 

If  all  are  taken  at  once,  the  work  should  be  carried  out  as  follows : 
The  numerator  of  the  sum  is 

(a;  _  3)(x  -  4)  +  (x  -  l)(a:  -  4)  +  (a:  -  l)(x  -  2). 

Adding  the  first  two  terms  with  respect  to  (x  —  4),  we  have 

2(a:-2)(a:-4)  +  (x  -  l)(x  -  2). 
Adding  these  with  respect  to  (x  —  2),  we  have  d(x  —  3)(x  —  2). 
Hence  the  sum  is  S(x-3)(x-2) 


(X  -  IXx  -  2){x  -  3)(x-  -  4)      ix-l)ix-  4) 
EXERCISES 

Perform  the  following  indicated  additions  and  subtractions : 
2,3  4  o3  5  1 


x-3     x-4:     x-5  4(a;H-3)     8(a?+5)    8(a;+l) 

4^7 


2(a;-l)      x-2     2(a;-3) 
1  7         .        13 


12(a;  +  l)     3(a;-2)      4(a;-3) 


404  ALGEBRAIC  FRACTIONS 

2  3            4  ^       6x^6  3a;-4 

•    (x-\-iy  x  +  1     x-^2'  '  ic^  +  aj  +  l  ic^-x  +  l 

1  x-2  „          1         .  4iB-8 


3(aj  +  l)      3(a^-a;  +  l)  5  (x  +  2)     5  {x' -{- 1) 

9.    ^-2__      1      +      1 


(a;  _  2)2     x-2     x-^1 


11. 


(x-2y     6(x-2)     Bix'-Jrl) 
1.1  1 


(x-iy      (x-l)      (a^-l) 


X3.  ,,,,J^,+.,.^A_  +  ,,,,A^+       1 


13. 


2(l-3a;)3    8(l-3a;)2     32(l-3a;)     32(l  +  a;) 

1 1 + ? 

(1  _  a)(2  -  a)      (2  -  a)(a  -  3)      (3  -  a)(a  -  1) 

ajy  yz  xz 


14. 

(2;  _  ^,)(aj  _  2!)      (x-z)(x-y)      (y-xXy-z) 

15        1  2a-5         5a^-Sa-2 

"  *   a  -  1      a2  -  2  a  +  1  (a  -  1)' 

16.   i ^^ +-1^?L±2_. 

m^  H-  m  +  1      m^  —  m  +  1     m'^  +  m^  +  1 

17.  1  ■  1  2 


62 -.364-26* -56  +  6     62__4^^3 

r  +  g 3  +  ^ r  +  ^ 

(r-t)(8-t)     (r-sXt-r)      (t  ~  s)(s -^  r) 


19  P'  +  ^'         _^        g^-pr        _^        T^-\-pq 

(p-9)(P  +  r)      (9-r){q-p)      (r-q)(r+p) 

^^  3aj2  4-l  20^2^2 


5ic2-18a;  +  9     4cc2_ii3,_3 


MULTIPLICATION  AND   DIVISION  OF  FRACTIONS     405 

MULTIPLICATION  AND  DIVISION  OP  PRACTIONS 

160.  The  product  of  two  fractions  is  a  fraction  whose  numera- 
tor is  the  product  of  the  given  numerators  and  whose  denomina- 
tor is  the  product  of  the  given  denominators. 


a     n  _an 
b  '  (fbd 


That  is, 

For  let  a  =  r  •  -3- 

0      a 

Then  hdx  =  bdl^-^\  §7 

§  8 
§11 


hdx  =  b'    ^    ' 

b 

"•i 

bdx  =  an. 

Hence, 

.  '=i- 

Therefore, 

a      n  _an 
b  '  d      bd' 

§2 

It  follows  that  a  frcxction  is  raised  to  any  power  by  raising 
numerator  and  denominator  separately  to  that  power. 


a      a      a-"     a      a     a      a 


For  -  .  "  =  i^,    ^  .  it  .  ^  =  ii-,  etc. 
b      b      b^'    b     b     b      68' 

A  fraction  multiplied  by  itself  inverted  equals  +  1. 

For  !?.^  =  !^  =  +  land  -~  .  (-i)=^  =  +  l. 
d     n      nd  d      \     nj      nd 

161.  Definitions.  If  the  product  of  two  numbers  is  1,  each 
is  called  the  reciprocal  of  the  other.  Hence  by  §  160,  the  re- 
ciprocal of  a  fraction  is  the  fraction  inverted. 

Also,  since  from  ab  =  l  we  have  a  =  -  and  &  =  -,  it  follows 

b  a 

that  if  two  numbers  are  reciprocals  of  each  other,  then  either 

one  is  the  quotient  obtained  by  dividing  1  by  the  other. 


406  ALGEBRAIC   FRACTIONS 

162.    To  divide  hy  any  nuuiber  is  equivalent  to  multi- 
plying hy  its  reciprocal. 

For  it  is  an  immediate  consequence  of  §  29  that 

n-7-d  or -=n'-' 
d  d 

To  divide  a  number  by  a  fraction  is  equivalent  to  multiplying 
by  the  fraction  inverted. 

For  by  §  161  the  reciprocal  of  the  fraction  is  the  fraction  inverted. 

A  fraction  is  divided  by  an  integer  by  multiplying  its  denomi- 
nator or  dividing  its  numerator  by  that  integer. 

For 


n 

d  '' 

n    1        n 
-a  =3'-  =—5 
d    a      ad 

n 

^a  =  ^-^«, since 
d 

and  rL^a  =  ""-^-^^  since  ^  =  ^L^tj^  by  §  47. 

•  d  d  ad         d 

In  multiplying  and  dividing  fractions  their  terms  should  at 
once  be  put  into  factored  forms. 

When  mixed  expressions  or  sums  of  fractions  are  to  be  mul- 
tiplied or  divided,  these  operations  are  indicated  by  means  of 
parentheses,  and  the  additions  or  subtractions  within  the 
parentheses  should  be  performed  first,  §  38. 

..s,„p>>„[(.-..j|i).(ji,-a,)].J^. 

Performing  the  indicated  operations  within  the  parentheses,  we  have 

Vl  +  a'^  ,     2  a-]       3a8   ^l  +  a^    a2_i  3  a«  ^      3  a^ 

Ll+a   ■  a2-lJ  *  a^-l     1  +  a*     2a     *  (a^ -r  l)(a2+l)     2(a  +  l)* 

EXERCISES 

Perform  the  following  indicated  operations  and  reduce  each. 
result  to  its  simplest  form. 

x^  +  x^y"^  -h  y^  ^  x"^  —  y^ 
a^  —  y^  a^-\-y^ 


MULTIPLICATION  AND   DIVISION   OF   FRACTIONS      407 

g^  —  6 V  4-  acy?  —  hax^    ,     —  a'}f  —  hxy^  —  cx^if 
'   36a*-9d'-\-24:a-16  '  20 x" -15 aaf - 30 d'a^' 

20  r^s^  +  23  rst  —  21 1^  12  mn^  -  28  mn^y  -  24  m/iy^ 

'   8  mhi^  -  48  mH^y  +  72  m'^^zy^         -j^q  ^^  ^  24  rsi  - 18  f 

\h.    a)\h^     a"     b     ~a       J  '   a-b'  L'  2s-\-3t^ 

\x     yj\x     yj\       x  +  yj\       x-yj 

V       a-b)\       a  +  bj     \       d'-by  ^_ 

\m  —  w     m  +  nj\  m  —  nj     \m  —  7i     m-\-nJ 

10.  ^^±^±i+_£L_^  .  (J^±yL\  .  fl+-^\ 

\    x  +  y    ^{x  +  yf)     \{x^yf-z'J     \   ^ x-\-yJ 

jj     a^  +  a^  +  6'      a  +  b      (  2     b^-a^b\ 
a2-a6  +  62  *  ^^-ft^  *  \^  "*"   a  +  6  / 

-„     m^  +  mn  ^  m^  —  mn^  —  m^ 4- n^     m^?i^  +  mn^  +  n^ 
m^  +  n^  m^n  —  w*  m*  —  2  m^  4-  m^ 


m^n  4-  2  m^n^  4-  mw 


18. 


(v+A-JgO.[S±^.§-l)-^.(i-i)} 


408  ALGEBRAIC   FRACTIONS 

COMPLEX  FRACTIONS 

163.  A  fraction  which  contains  a  fraction  either  in  its 
numerator  or  in  its  denominator  or  in  both  is  called  a  complex 
fraction.    • 

Since  every  fraction  is  an  indicated  operation  in  division, 
any  complex  fraction  may  be  simplified  by  performing  the  indi- 
cated division. 

It  is  usually  better,  however,  to  remove  all  the  minor 
denominators  at  once  by  multiplying  both  terms  of  the  com- 
plex fraction  by  the  least  common  multiple  of  all  the  minor 
denominators  according  to  §  47. 

mle        3     2    _     V3     2/        _2x  +  ^x 


For  example,     J^^  .^\  ''".       ^2^  +  Sx^_5^ 
3       2 


2x2     3      /2^_3\g      4a:2-.9      4a;2-9 


A  complex  fraction  may  contain  another  complex  fraction 
in  one  of  its  terms. 

E.a.     1 —   has  the  complex  fraction  = — 


a  —  1 

in  its  denominator.    This  latter  fraction  is  first  reduced  by  multiply- 
ing its  numerator  and  denominator  by  a  —  1,  giving 

q  +  l        _  q2-l 

1         a2-a  +  l* 


a  + 


a-1 


Substituting  this  result  in  the  given  fraction,  we  have 

1  ^  1  ^flg-g  +  l 

\^     1  ^a^-a  +  1 

a  + T 

a  —  1 


COMPLEX  FRACTIONS  '  409 


EXERCISES 

Simplify  each  of  the  following, 


2. 


4. 


6. 


m  —  nm  +  n  y^ 


-h 

a-2-\-- 
a 

a^  +  ab 
a-b 

a-2  +  ?-i 
a     a? 

1 

^y  —  y* 

'-.!. 

11      .2a 


a  +  x     a  —  x     a^—x^ 
~T  1  2^ 


a  +  x     a  —  x     (J?  — 7?  1  — 

1      .      1      .      2a 


4-^-  +  -^:^  1_i 


a  +  a;     a  — a:     a^+ic^ 
~1  1  2a; 


1-i 


a  — a;     a-\-x     (v--\-7? 

,3         «8 


10. 


m2-mn  +  n2_^Lz_IL  3  + 


m  +  w  Q  j_  ^ 


mr -\- mn -\- n -\ ■ —  6-\-- 

m  — 71  X 

EQUATIONS  INVOLVING   ALGEBRAIC  FRACTIONS 
164.   In  solving  a  fractional  equation,  it  is  usually  conven- 
ient to  clear  it  of  fractions,  that  is,  to  transform  it  into  an  equiv- 
alent equation  containing  no  fractions. 

In  case  no  denominator  contains  any  unknown  this  may  be 
done  by  multiplying  both  members  by  the  L.  C.  M.  of  all  the 
denominators,  §  62. 


410  ALGEBRAIC  FRACTIONS 

When,  however,  the  unknown  appears  in  any  denominator, 
multiplying  by  the  L.  C.  M.  of  all  the  denominators  may  or  may 
not  introduce  new  roots,  as  shown  in  the  following  examples. 

It  may  easily  be  shown,  that  multiplying  an  integral  equation 
by  any  expression  containing  the  unknown  always  introduces  new 
roots. 

Ex.1.    Solve  _i^  +  _J_  =  2.  (1) 

x—2     x—S 

Clearing  of  fractions  by  multiplying  by  (x  —  2)  (a:  —  3),  and  simpli- 
fying,wehave    2x^  _  13^  +  ,0  =(x -4)(2x  -  5)=  0.  (2) 

The  roots  of  (2)  are  4  and  2|,  both  of  which  satisfy  (1).  Hence 
no  new  root  was  introduced  by  clearing  of  fractions. 

Ex.2.   Solve         ^-(,_4_,)-  (1) 

Clearing  of  fraction  a>  we  have, 

a;  -  2  =  1.  (2) 

The  only  root  of  (2)  is  z  =  3,  which  is  also  the  only  root  of  (1). 
Hence  no  new  root  was  introduced. 

Ex.3.   Solve  .^11^  =  1.  (1) 

ar  — 4 

Clearing  of  fractions  and  simplifying,  we  have, 

x-^  -  X  -  2  =  (X  -  2)(x  +  1)  =  0.  (2) 

The  roots  of  (2)  are  2  and  —  1.  Now  x  =  —  1  is  a  root  of  (1),  but 
X  =  2  is  not,  since  we  are  not  permitted  to  make  a  substitution  which 
reduces  a  denominator  to  zero,  §  50.  Hence  a  new  root  has  been  in- 
troduced and  (1)  and  (2)  are  not  equivalent. 

x-2 
If  the  fraction    2_  4  ^  ^^^t  reduced  to  lowest  terms,  we  have  the 

equation  1 

x  +  2  ^  \ 

Clearing  of  fractions.  a:  +  2  =  1.  (4) 

Now  (3)  and  (4)  are  equivalent,  —  1  being  the  only  root  of  each. 


EQUATIONS   INVOLVING  FRACTIONS  411 

•    Ex.4.   Solve  ^£^-^  =  1.  (1) 

ar—  1     x  —  1  ^  ^ 

Clearing  of  fractions  and  simplifying, 

x2-a:  =  ar(a:- 1)  =  0.  (2) 

The  roots  of  (2)  are  0  and  1.  x  =  0  satisfies  (1),  but  a:  =  1  does 
not,  since  it  is  not  a  permissible  substitution  in  either  fraction  of  (1). 
Hence  a  new  root  has  been  introduced. 

165.  Examples  3  and  4  illustrate  the  only  cases  in  which  new 
roots  can  be  iritroduced  by  multiplying  by  the  L.  C.  M.  of  the 
denominators. 

This  can  be  shown  by  proving  certain  theorems,  the  results 
of  which  are  here  used  in  the  following  directions  for  solving 
fractional  equations : 

(1)  Eeduce  all  fractions  to  their  lowest  terms. 

(2)  Multiply  both  members  by  the  least  common  multiple 
of  the  denominators. 

(3)  Reject  any  root  of  the  resulting  equation  which  reduces 
any  denominator  of  the  given  equation  to  zero.  The  remain- 
ing roots  will  then  satisfy  both  equations,  and  hence  are  the 
solutions  desired. 

If  when  each  fraction  is  in  its  lowest  terms  the  given  equation  con- 
tains no  two  which  have  a  factor  common  to  their  denominators,  then 
no  new  root  can  enter  the  resulting  equation  and  none  need  to  be  re- 
jected.    See  Ex.  1  and  Ex.  3  after  being  reduced. 

If,  however,  any  two  or  more  denominators  have  some  common 
factor  X  —  a,  then  x  =  a  may  or  may  not  be  a  new  root  in  the  resulting 
equation,  but  in  any  case  it  is  the  only  possible  new  root  which  can 
enter,  and  must  be  tested.     Compare  Exs.  2  and  4. 

Ex.5.    Examine  _^^±^- +  _A^-_ - ^  =  8. 

Since  each  fraction  is  in  its  lowest  terms  and  no  two  dignominators 
contain  a  common  factor,  then  clearing  of  fractions  will  give  an  equa- 
tion equivalent  to  the  given  one? 


412  ALGEBRAIC  FRACTIONS 

Ex.  6.    Examine  -/^t3     __       ^-7      ^ 4^ 

Each  fraction  is  in  its  lowest  terms,  but  the  two  denominators  have 
the  factor  a:  +  2  in  common.  Hence  x  =  —  2  is  the  only  possible  new 
root  which  can  enter  the  resulting  integral  equation,  but  on  trial  it  is 
found  not  to  be  a  root.     Hence  the  two  equations  are  equivalent. 

EXERCISES 

Determine  whether  each  of  the  following  when  cleared  of 
fractions  produces  an  equivalent  equation,  and  solve  each. 

Sar'-Ta  +  S  3^-4 


*    2x-l     x+1  x-1 

3(a;-l)      x'-l     4  b  a 

6.    2a-l_^l^     3a    ^  ^^     6  =  .^z:^. 
a          23a  — 1  1  —  ax 

-        2     +^  =  -±-.  11.    _2      ^,0^^^     2 


ic  — aa;  — 6      a;  — c  a;  — 10  10  — a? 

g     _1 L_  =  _3±^.    12.    §z:^4.^i=i  =  ^. 

*a  — a;     a  +  a;         o^  —  y?  *a;  — 4     6  — a;     2 

13  g       ,       24      ^2(a-4)     1 

*  2a-l"^4a2-l       2a4-l       9 

14  g      ,  g  — l__a^  +  <^  — 1^ 

*  g  — 1         g  a^— a 

15.    -J^ 5-^  +  ^^ 

a*-4     2-a^^     3(a  +  2) 


17. 


EQUATIONS  INVOLVING  FRACTIONS  413 

\ 
,_     ax-^b  '.  cx-\-d     ax  —  b.cx  —  d 
a-\-bx     c-\-dx     a  —  bx     c  — dx 

(ft  —  x)(x  —  b)    _  -g     x-\-m,  —  2n_n-\-2m  —  2x 

(a  —  x)  —  {x  —  b)  x-{-m  +  2n     n  —  2m-{-2x 

19  (a-xy-(x-by^   4a6 

(a  —  x){x—  b)        a'  —  b^ 

20  l±Sx_9-llx  _(2x-Sl 
'    5  +  7a;      5_7a.  26-49a!« 

_,      a;-f-2a  ,  a;  — 2a_     4a& 


26-a;     26  +  a;    ,462-0^ 


22.    _1^+        ^^  ' 


-2     24(a;4-2)     ar^_4 

23.  ^  +  «  I  a?  — a_      2(a^  +  l) 

a;  —  ft     a;  +  ft      (1  -|-  ft)  (1  —  a) 

24.  ^n25  =  rL^.  25.    _A__    2a    ^j,_ 
a;^m     n  +  x  a;  — ft     xr  —  a^ 

26.    -^-  +  ^(^^-^)  =  ^£±l. 
3a;H-l        2a;  +  l        3a;4-l 

2a;  +  3        7-3a;        a;-7    ^^ 
*    2(2a;-l)'^3a;-4'^2(a;  +  l)        * 

1      , ft— 6         1      , a+b 

za.     ri =  — —r-\ • 

ft  — 0         X         ft  +  6         a; 

1  p 


29. 


3  (?M  +  ny     m-\-n     2  (m  +  n) 


p'^a;  jp 


30    y'4-2y-2       y    ^    y 


414  ALGEBRAIC   FRACTIONS 

31  ^^  7      ^8a;^-13a;-64 

*    2x-\-3     3x-4:        6a^  +  x-12    ' 


ir  —  a^     ic  —  a     x^  +  ax  -\-a^ 


33. 


l_2a;     5-6a;     8  l-Sa^^ 


3_4a;     7-8a;     3    21  -  52  a;  +  32  a;^ 


34.    m-g  ^  n-p^m-q  ^  n-p 
x—n      x—q      x—p     x—m 


•    x-3     x-2     4:x'-20x  +  24:       ' 


36.    -^ ? +  -l-  =  0. 

a^-\-27     a^-Sx-\-d     x  +  3 

x—9     x—7     x—9     x—S     x—7     x—S 


37, 


x  —  5     x  —  2     ic  — 4     x—5     x  —  4:     x  —  2 


38.    o^(^  +  ^-c)(^-^  +  c) 
(6  +  c  +  a?)  (6  +  c  —  ») 


PROBLEMS 

1.  Find  a  number  such  that  if  it  is  added  to  each  term  of 
the  fractiou  |  and  subtracted  from  each  term  of  the  fraction 
i|  the  results  will  be  equal. 

2.  Make  and  solve  a  general  problem  of  which  1  is  a 
special  case. 

3.  Three  times  one  of  two  numbers  is  4  times  the  other. 
If  the  sum  of  their  squares  is  divided  by  the  sum  of  the  num- 
bers, the  quotient  is  42^  less  than  that  obtained  by  dividing 
the  sum  of  the  squares  by  the  difference  of  the  numbers.  Find 
the  numbers. 


PROBLEMS  415 

4.  The  sum  of  two  numbers  less  2,  divided  by  their  differ- 
ence, is  4,  and  the  sum  of  their  cubes  divided  by  the  difference 
of  their  squares  is  If  times  their  sum.     Find  the  numbers. 

5.  The  circumference  of  the  rear  wheel  of  a  carriage  is 
4  feet  greater  than  that  of  the  front  wheel.*  In  running  one 
mile  the  front  wheel  makes  110  revolutions  more  than  the  rear 
wheel.     Find  the  circumference  of  each  wheel. 

6.  State  and  solve  a  general  problem  of  which  5  is  a  special 
case,  using  b  feet  instead  of  one  mile,  letting  the  other  num- 
bers remain  as  they  are  in  problem  5. 

7.  In  going  one  mile  the  front  wheel  of  a  carriage  makes 
88  revolutions  more  than  the  rear  wheel.  If  one  foot  is  added 
to  the  circumference  of  the  rear  wheel,  and  3  feet  to  that  of 
the  front  wheel,  the  latter  will  make  22  revolutions  more  than 
the  former.     Find  the  circumference  of  each  wheel. 

8.  State  and  solve  a  general  problem  of  which  5  is  a  special 
case,  using  letters  for  all  the  numbers  involved. 

9.  The  circumference  of  the  front  wheel  of  a  carriage  is 
a  feet,  and  that  of  the  rear  wheel  b  feet.  In  going  a  certain 
distance  the  front  wheel  makes  n  revolutions  more  than  the 
rear  wheel.     Find  the  distance. 

10.  State  and  solve  a  problem  which  is  a  special  case  of 
problem  9,  using  the  formula  just  obtained. 

11.  Find  a  number  consisting  of  two  digits  whose  sum, 
divided  by  their  difference,  is  4.  This  number  divided  by  the 
sum  of  its  digits  is  equal  to  twice  the  digit  in  units'  place 
plus  \  of  the  digit  in  tens'  place,  the  tens'  digit  being  the 
greater. 

12.  There  is  a  fraction  such  that  if  3  is  added  to  each  of  its 
terms,  the  result  is  ^,  and  if  3  is  subtracted  from  each  of  its 
terms,  the  result  is  ^.     Find  the  fraction. 

13.  State  and  solve  a  general  problem  of  which  12  is  a 
special  case. 


416  ALGEBRAIC  FRACTIONS 

14.  A  and  B  working  together  can  do  a  piece  of  work  in 
6  days.  A  can  do  it  alone  in  5  days  less  than  B.  How  long 
will  it  require  each  when  working  alone  ? 

15.  State  and  solve  a  general  problem  of  which  14  is  a 
special  case. 

16.  On  her  second  westward  trip  the  Mauretania  traveled 
625  knots  in  a  certain  time.  If  her  speed  had  been  5  knots 
less  per  hour,  it  would  have  required  Q^  hours'  longer  to  cover 
the  same  distance.     Find  her  speed  per  hour. 

17.  By  increasing  the  speed  a  miles  per  hour,  it  requires 
h  hours  less  to  go  c  miles.  Eind  the  original  speed.  Show 
how  problem  16  may  be  solved  by  means  of  the  formula  thus 
obtained. 

18.  A  train  is  to  run  d  miles  in  a  hours.  After  going  c  miles 
a  dispatch  is  received  requiring  the  train  to  reach  its  destina- 
tion h  hours  earlier.  What  must  be  the  speed  of  the  train  for 
the  remainder  of  the  journey  ? 

19.  A  man  can  row  a  miles  down  stream  and  return  in  h 
hours.  If  his  rate  up  stream  is  c  miles  per  hour  less  than 
down  stream,  find  the  rate  of  the  current,  and  the  rate  of  the 
boat  in  still  water. 

20.  State  and  solve  a  special  case  of  problem  19. 

21.  A  can  do  a  piece  of  work  in  a  days,  B  can  do  it  in  h  days, 
and  C  in  c  days.  How  long  will  it  require  all  working  to- 
gether to  do  it  ? 

22.  Three  partners.  A,  B,  and  C,  are  to  divide  a  profit  of  p 
dollars.  A  had  put  in  a  dollars  for  m  months,  B  had  put  in 
h  dollars  for  n  months,  and  C  c  dollars  for  t  months.  What 
share  of  the  profit  does  each  get  ? 

23.  State  and  solve  a  problem  which  is  a  special  case  of  the 
preceding  problem. 


CHAPTER  IX 

RATIO,  VARIATION,  AND  PROPORTION 

RATIO  AND  VARIATION 

166.  In  many  important  applications  fractions  are  called 
ratios. 

E.g.    I  is  called  the  ratio  of  3  to  5  and  is  sometimes  -written  3  :  5. 

It  is  to  be  understood  that  a  ratio  is  the  quotient  of  two  num- 
bers and  hence  is  itself  a  number.  We  sometimes  speak  of  the 
ratio  of  two  magnitudes  of  the  same  kind,  meaning  thereby 
that  these  magnitudes  are  expressed  in  terms  of  a  common  unit 
and  a  ratio  formed  from  the  resulting  numbers. 

E.g.  If,  on  measuring,  the  heights  of  two  trees  are  found  to  be  25 
feet  and  35  feet  respectively,  we  say  the  ratio  of  their  heights  is  ||  or  ^. 

167.  Two  magnitudes  are  said  to  be  incommensurable  if  there 
is  no  common  unit  of  measure  which  is  contained  exactly  an 
integral  number  of  times  in  each. 

E.g.    If  a  and  d  are  the  lengths  of  the  side  and  the  diagonal  of  a 

square,  then'  d^  =  a"^  +  a^  §  166,  E.  C.     Hence,  ^=  1  or  -  =  i    But 

since  V2  is  neither  an  integer  nor  a  fraction  (§  108),  it  follows  that  a 
and  d  have  no  common  measure,  that  is,  they  are  incommensurable. 

168.  In  many  problems,  especially  in  Physics,  magnitudes 
are  considered  which  are  constantly  changing.  Number  ex- 
pressions representing  such  magnitudes  are  called  variables, 
while  those  which  represent  fixed  magnitudes  are  constants. 

E.g.  Suppose  a  body  is  moving  at  a  uniform  rate  of  5  ft.  per 
second.  If  <  is  the  number  of  seconds  from  the  time  of  starting 
and  s  the  number  of  feet  passed  over,  then  s  and  t  are  variables. 

417 


418  RATIO,   VARIATION,   AND  PROPORTION 

The  variables  s  and  t,  in  case  of  ur.iforra  motion,  have  a  Jixed  ratio; 
namely,  in  this  example,  5:^=5  for  every  pair  of  corresponding 
values  of  s  and  t  throughout  the  period  of  motion. 

169.  When  two  variables  are  so  related  that  for  all  pairs  of 
corresponding  values,  their  ratio  remains  constant^  then  each 
one  is  said  to  vary  directly  as  the  other. 

E.g.  If  s  :  f  =  A:  (a  constant)  then  s  varies  directly  as  «,  and  t  varies 
directly  as  s. 

Variation  is  sometimes  indicated  by  the  symbol  oc.     Thus 

s  oc  ^  means  s  varies  as  t,\.Q.  -  =  h  ov  s  =  Jet. 

t 

170.  When  two  variables  are  so  related  that  for  all  pairs  of 
corresponding  values  their  product  remains  constant,  then  each 
one  is  said  to  vary  inversely  as  the  other. 

E.g.  Consider  a  rectangle  whose  area  is  A  and  whose  base  and 
altitude  are  I  and  h  respectively.  '  Then,  A  =h  -h. 

If  now  the  base  is  multiplied  by  2,  3,  4,  etc.,  while  the  altitude  is 
divided  by  2,  3,  4,  etc.,  then  the  area  will  remain  constant.  Hence, 
h  and  h  may  both  vary  while  A  remains  constant. 

The  relation  b  -h  =  A  may  be  written  b  =  A  •  -  ov  h  =  A  '  —     It 

11^. 
may  also  be  written  b:-  =  A   or  h  :  ~  =  A,  so  that  the  ratio  of  either 

h  b 

b  or  hto  the  reciprocal  of  the  other  is  the  constant  A.     For  this  rea- 
son one  is  said  to  vai'y  inversely  as  the  other. 

171.  If  2/  =  ^^^  ^  being  constant  and  x  and  y  variables,  then 
y  varies  directly  as  x^.  li  y  =  —,  then  y  varies  inversely  as  x^. 
liy  =Jc  -  wx,  then  y  varies  jointly  as  w  and  x.  li  y  cc  wx,  then  y 
oc  w  if  a;  is  constant  and y  ccxiiwi^  constant,   liy^h  •  — ,  then 

X 

y  varies  directly  as  w  and  inversely  as  x. 

Example.  The  resistance  offered  by  a  wire  to  an  electric 
current  varies  directly  as  its  length  and  inversely  as  the  area 
of  its  cross  section. 

If  a  wire  \  in.  in  diameter  has  a  resistance  of  r  units  per  mile, 
find  the  resistance  of  a  wire  \  in.  in  diameter  and  3  miles  long. 


RATIO   AND  VARIATION  '  419 

Solution.     Let  R  represent  the  resistance  of  a  wire  of  length  I  and 

cross-section  area  s  —ir  •  (radius)^.     Then  R  =  k  •  -  where  k  is  some 

s 
constant.        Since    R  =  r   when  1  =  1    and    s  =  7r(xV)^    we    have 

^  _  z; .  J_  or  ^•  =  -^ 
'^-^    _7r_  256 

256 
Hence,  when  I  =  S  and  s  =  irdy,  we  have, 

256    jr       4 
64 

That  is,  the  resistance  of  three  miles  of  the  second  wire  is 
j  the  resistance  ^^er  mile  of  the  first  wire. 

PROBLEMS 

1.  If  z  oc  Wj  and  if  2  =  27  when  w  =  3,  find  the  value  of  z 
when  to  =  4J. 

2.  If  z  varies  jointly  as  to  and  x,  and  if  2  =  24  when  w  =  2 
and  a;  =  3,  find  z  when  t«  =  3^  and  x  =  7. 

3.  If  2  varies  inversely  as  w,  and  if  z  =  ll  when  t^  =  3, 
find  2  when  to  =  66. 

4.  If  2  varies  directly  as  iv  and  inversely  as  cc,  and  if  2;  =  28 
when  w  =  14  and  a;  =  2,  find  z  when  m;  =  42  and  a;  =  3. 

5.  If  z  varies  inversely  as  the  square  of  w,  and  if  2  =  3 
when  w  =  2,  find  2  when  w  =  6. 

6.  If  q  varies  directly  as  m  and  inversely  as  the  square 
of  d,  and  q  =  30  when  m  =  1  and  d  ==  y|-Q^,  find  q  when  m  =  3 
and  d  =  600. 

7.  If  y^  oc  a:^,  and  if  y  =  16  when  aj  =  4,  find  y  when  a;  =  9. 

8.  The  weight  of  a  triangle  cut  from  a  steel  plate  of  uni- 
form thickness  varies  jointly  as  its  base  and  altitude.  Find 
the  base  when  the  altitude  is  4  and  the  weight  72,  if  it  is 
known  that  the  weight  is  60  when  the  altitude  is  5  and  base  6. 


420  EATIO,  •  VARIATION,   AND   PROPORTION 

9.  The  weight  of  a  circular  piece  of  steel  cut  from  a  sheet 
of  uniform  thickness  varies  as  the  square  of  its  radius.  Find 
the  weight  of  a  piece  whose  radius  is"  13  ft.,  if  a  piece  of 
radius  7  feet  weighs  196  pounds. 

10.  If  a  body  starts  falling  from  rest,  its  velocity  varies 
directly  as  the  number  of  seconds  during  which  it  has  fallen. 
If  the  velocity  at  the  end  of  3  seconds  is  96.6  feet  per  second, 
find  its  velocity  at  the  end  of  7  seconds ;  of  ten  seconds. 

11.  If  a  body  starts  falling  from  rest,  the  total  distance 
fallen  varies  directly  as  the  square  of  the  time  during  which  it 
has  fallen.  If  in  2  seconds  it  falls  64.4  feet,  how  far  will  it 
fall  in  5  seconds  ?     In  9  seconds  ? 

12.  The  number  of  vibrations  per  second  of  a  pendulum 
varies  inversely  as  the  square  root  of  the  length.  If  a  pen- 
dulum 39.1  inches  long  vibrates  once  in  each  second,  how 
long  is  a  pendulum  which  vibrates  3  times  in  each  second? 

13.  Illuminating  gas  in  cities  is  forced  through  the  pipes  by 
subjecting  it  to  pressure  in  the  storage  tanks.  It  is  found  that 
the  volume  of  gas  varies  inversely  as  the  pressu;:e.  A  certain 
body  of  gas  occupies  49,000  cu.  ft.  when  under  a  pressure  of  2 
pounds  per  square  inch.  What  space  would  it  occupy  under 
a  pressure  of  2^  pounds  per  square  inch  ? 

14.  The  amount  of  heat  received  from  a  stove  varies  in- 
versely as  the  square  of  the  distance  from  it.  A  person  sitting 
15  feet  from  the  stove  moves  up  to  5  feet  from  it.  How  much 
will  this  increase  the  amount  of  heat  received  ? 

15.  The  weights  of  bodies  of  the  same  shape  and  of  the  same 
material  vary  as  the  cubes  of  corresponding  dimensions.  If  a 
ball  3;^  inches  in  diameter  weighs  14  oz.,  how  much  will  a  ball 
of  the  same  material  weigh  whose  diameter  is  3^^^  inches  ? 

16.  On  the  principle  of  problem  15,  if  a  man  5  feet  9  inches 
tall  weighs  165  pounds,  what  should  be  the  weight  of  a  man  of 
similar  build  6  feet  tall  ? 


PROPORTION  421 


PROPORTION 

172.  Definitions.  The  four  numbers  a,  6,  c,  d  are  said  to  be 
proportional  or  to  form  a  proportion  if  the  ratio  of  a  to  6  is  equal 

to  the  ratio  of  c  to  d.     That  is,  if  -=  -•   This  is  also  sometimes 

b     d 

written  a:b::c:d,  and  is  read  a  is  to  6  as  c  is  to  d. 

The  four  numbers  are  called  the  terms  of  the  proportion; 
the  first  and  fourth  are  the  extremes ;  the  second  and  third  the 
means  of  the  proportion.  The  first  and  third  are  the  antece- 
dents of  the  ratios,  the  second  and  fourth  the  consequents. 

If  a,  by  c,  X  are  proportional,  x  is  called  the  fourth  propor- 
tional to  a,  b,  c.  If  a,  x,  x,  b  are  proportional,  x  is  called  a  mean 
proportional  to  a  and  b,  and  b  a  third  proportional  to  a  and  x. 

173.  If  four  numbers  are  proportional  when  taken  in  a  given 
order,  there  are  other  orders  in  which  they  are  also  proportional. 

E.g.  If  a,  b,  c,  d  are  proportional  in  this  order,  they  are  also  pro- 
portional in  the  following  orders:  a,  c,  5,  d;  h,  a,  d,  c;  b,  d,  a,  c; 
c,  a,  rf,  h ;  c,  d,  a,  b ;  d,  c,  b,  a  ;  and  d,  b,  c,  a. 

Ex.  1.  Write  in  the  form  of  an  equation  the  proportion  cor- 
responding to  each  set  of  four  numbers  given  above,  and  show 

how  each  may  be  derived  from  -  =  -.     See  §  196,  E.  C. 

b     d 

Show  first  how  to  derive  -  =  -  (1),  and  then  -  =  -  (2). 
c     d      ^  a     c  ^  ^ 

Derive  also       «±^  =  £+_?[(3)  ^nd  ^^^  =  ^^(4). 
a  c^^  a  c     ^ 

In  (1)  the  original  proportion  is  said  to  be  taken  by  alterna- 
tion, and  in  (2)  by  inversion ;  in  (3)  by  composition,  and  in  (4) 
by  division. 

Ex.  2.   From  -  =  -  and  (1),  (2),  (3),  (4)  obtain  the  following. 

0       Ui 

See  pp.  279-281,  E.  C. 

a±b  __c  ±d  a±h  _a  a±c  _a  a±b  _a~b 

b  d    ^  c±d~  c'  b±d~  b'  c  +  rf~cTc?* 

When  the  double  sign  occurs,  the  upper  signs  are  to  be  read  together 
and  the  lower  signs  together. 


422  RATIO,    VARIATION,   AND   PROPORTION 

EXERCISES 

1.  What  principles  in  the  transformation  of  equations  are 
involved  (a)  in  taking  a  proportion  by  inversion,  (b)  by  alter- 
nation, (c)  by  composition,  (c^)  by  division,  (e)  by  composition 
and  division  ? 

2.  From  ^  =  ^  derive  ^-i—  =  -  and  state  all  the  principles 
involved.       *      '^  a  +  h      b 

3.  From  a:b::c:d  show  that  a^ :  b^ : :  c^ : d^. 

4.  From  a:b::c:d  and  m:n::r:s  show  that  am:bn::cr:  ds. 

5.  From  ^i  =  ^== ...  =^  show  that 

%  +  Ct2  +    •  •  •   +  ^w  _  <^1_  ^2  __    ...   _^ 

Suggestion.    Let  ^  =  ^  =  ...  =  ^  =  A:. 

^    Tjj  a     c!    1        ,-•    .  ma  —  b      a 

6.  If  -  =  -'  show  that =  -. 

b      d  mc~d     c 

7.  If  2  =  ^  show  that  ^±^  =  5+^. 

y     w  x  —  kyz  —  kw 

8.  If  H  =  ^  show  that  rm  +  nb^m^  +  nd 

b      d  ma  —  nb     mc—nd 

9.  (a)  Find  a  fourth  proportional  to  17,  19,  and  187. 
(6)  Find  a  mean  proportional  between  6  and  54. 
(c)  Find  a  third  proportional  to  27  and  189. 

10.   Find  the  unknown  term  in  each  of  the  following  pro 
portions : 

aj:42::27:126;  78:a;::13:3j  99: 117:  :a;:39;  171:27:  :57:aj. 


PROPORTION  423 

11.  If  Si  and  «2  are  two  distances  passed  over  by  a  body 

S  /  2 

falling  from  rest  in  the  time  intervals  t^  and  ^g?  then  —  =  — 

(see  jjroblem  11,  p.  420).     If  it  is  known  that  a  body  falls 
144.9  feet  in  3  seconds,  how  far  will  it  fall  in  8  seconds  ? 

12.  When  a  weight  is  attached  to  a  spring  balance  the  in- 
dex is  displaced  a  distance  proportional  to  the  weight.  Thus, 
if  di  and  c?2  are  displacements  and  Wi,  w.2  the  corresponding 

weights,  then  -^=  ^.     If  a  2-pound  weight  displaces  the  in- 

c?2     ^2 
dex  \  inch,  how  much  will  a  50-pound  weight  displace  it  ? 

13.  The  intensity  of  light  is  inversely  proportional  to  the 

square  of  the  distance  from  the  source  of  the  light.    That  is,  if 

t'l  and  12  are  the  measures  of  intensities  at  the  distances  d^  and 

i      d^ 
dzj  then  -^  =  -^,.    If  the  intensity  of  a  given  light  at  a  distance 

^2         ^X 

of  2  feet  is  20  candle  power,  find  the  intensity  at  5  feet  ? 

14.  If  iv^  and  W2  are  weights  resting  on  the  two  ends  of  a 
beam,  and  if  the  distances  from  the  fulcrum  are  d^  and  do  re- 
spectively, then  the  beam  will  balance  when  ^  =  -^ .    That  is, 

W2     d^ 

the  weights  are  inversely  proportional  to  the  distances. 

If  a  stone  weighing  850  pounds  at  a  distance  of  1  foot  from 
the  fulcrum  is  to  be  balanced  by  a  50-pound  weight,  where 
should  the  weight  be  applied  ? 

15.  Find  where  the  fulcrum  should  be  in  order  that  two 
boys  weighing  110  and  80  pounds  respectively  may  balance  on 
the  ends  of  a  16-foot  plank. 

16.  The  weight  of  a  body  above  the  earth's  surface  varies 
inversely  as  the  square  of  its  distance  from  the  earth's  center. 
If  an  object  weighs  2000  pounds  at  the  earth's  surface,  what 
would  be  its  weight  if  it  were  12,000  miles  above  the  center  of 
the  earth,  the  radius  of  the  earth  being  4000  miles  ? 


CHAPTER  X 
EXPONENTS  AND  RADICALS 

FRACTIONAL   AND   NEGATIVE   EXPONENTS 

174.  The  meaning  heretofore  attached  to  the  word  exponent 
cannot  apply  to  a  fractional  or  negative  number. 

E.g.  Such  an  exponent  as  f  or  —  5  cannot  indicate  the  number  of 
times  a  base  is  used  as  a  factor. 

It  is  possible,  however,  to  interpret  fractional  and  negative 
exponents  in  such  a  way  tJiat  the  laws  of  operations  which  gov- 
ern positive  integral  exponents  shall  apply  to  these  also. 

175.  The  laws  for  positive  integral  exponents  are : 

I.  a'"  '  a"  =  a"'+n,  §  43 

'                           II.  fl*"  -^  a"  =  a'"-".  §  46 

III.  {a"')"  =  a""".  §  115 

IV.  (a'»  '  b")P  =  a'"Pb"P,  §  116 
V.  (a"'  -^  s")P  =  a'"P  -^  s"p.  §  117 

176.  Assuming  Law  I  to  hold  for  positive  fractional  expo- 
nents and  letting  r  and  s  be  positive  integers,  we  determine  as 

r 

follows  the  meaning  of  b'  (read  b  exponent  r  divided  by  s). 

.       ( -Y     -    '- 
By  definition,  \if  J  z=  b*' b*  >"  to  8  factors, 

which  by  Law  I  =  jl'*"^''"  "•  **"  *«""«_  ^J  *  •  _  jr^ 

r 

Hence,  6»  is  one  of  the  s  equal  factors  of  6*". 

r  I 

That  is,  6»  =  v^,  and  in  particular  6»  =  </? .  See  §  114 

424 


FRACTIONAL  AND  NEGATIVE   EXPONENTS  425 

/  IV    1    1  r 

Similarly,  from  \b*  j  =  b' .b* --to  r  factors,  =  &•, 
we  show  that  6^  =  \b')  =  i</b  )\ 

Hence,  f»=W=(Vby.  See  §  119 

Thus  a  positive  fractional  exponent  means  a  root  of  a  power 
or  a  poiver  of  a  root,  the  numerator  indicating  the  power  and  the 
denominator  indicating  the  root. 

E.g.     a^  =  v^  =  ( v^  )=2;  8^  =  \^  =  4,  or  (\/8  y  =  2'^  =  4. 

177.  Assuming  Law  I  to  hold  also  for  negative  exponents, 
and  letting  ^  be  a  positive  number,  integral  or  fractional,  we 
determine  as  follows  the  meaning  of  6"'  {read  b  exponent  nega- 
tive t). 

By  Law  I,  5* .  6-«  =  fto  =  1.  §  46 

Therefore,  b-'  =  i.  §  11 

Hence  a  number  with  a  negative  exponent  means  the  same 
as  the  reciprocal  of  the  number  with  a  positive  exponent  of  the 
same  absolute  value. 

E.g.    a-  =  i.;        4-i=l  =  i  =  l. 
"  a'  ^1     2»     8 

* 

178.  It  thus  appears  that  fractional  and  negative  exponents 
simply  provide  new  ways  of  indicating  operations  already  well 
known.  Sometimes  one  notation  is  more  convenient  and  some- 
times the  other. 

^Fractional  and  negative  exponents  are  also  called  powers. 

E.g.  a;'  may  be  read  x  to  the  f  power,  and  a;-*  may  be  read  x  to  the 
—  4/A  power. 

The  limitations  as  to  principal  roots  and  the  sign  of  the  base, 
imposed  in  theorems  on  powers  and  roots  in  Chapter  VI,  neces- 
sarily apply  to  the  corresponding  theorems  in  this  chapter.  See 
§§  114-122. 


426  EXPONENTS  AND   KADICALS 

In  any  algebraic  expression,  radical  signs  may  now  be  re- 
placed by  fractional  exponents,  or  fractional  exponents  by 
radical  signs. 

In  a  fraction,  any  factor  may  be  changed  from  numerator  to 
denominator,  or  from  denominator  to  numerator,  by  changing 
the  sign  of  its  exponent. 

Ex.  1.    v^  +  3  v^  .  V^  +  5  V^y/f  =  x^  +  3 x^y^  +  5^^. 

Ex.  2.    —  =  ahx-^,  since  ahx-'^  =  ab--=  —  ' 
x'^  x^     x^ 

Ex.3.    ab-^c^  =  ac^'L  =  ^. 
b^       63 

Ex.4.   32-^--L-_^-l_l- 


32 


t      (</32y     24      16 


EXERCISES 


(a)  In  the  expressions  containing  radicals  on  p.  436,  replace  these 
by  fractional  exponents. 

(b)  Replace  all  positive  fractional  exponents  on  this  page  by 
radicals. 

(c)  Change  all  expressions  containing  negative  exponents  to  equiv- 
alent expressions  having  only  positive  exponents. 

179.  Fractional  and  negative  exponents  have  been  defined 
so  as  to  conform  to  Law  I,  §§  176,  177.  We  now  show  that 
when  so  defined  they  also  conform  to  Laws  II,  III,  IV,  and  V. 

To  verify  Law  II.  Since  by  Law  T,  a*"-"  •  a"  =  a^,  for  m  and  n  inte- 
gral or  fractional,  positive  or  negative,  it  follows  by  §  11  that 
a"*  -i-  a*^  =  a*"""  for  all  rational  exponents. 

To  verify  Law  III.  Let  r  and  s  be  positive  integers,  and  let  k  be 
any  positive  or  negative  integer  or  fraction.     Then  we  have : 

(1)  (a''y=  </(^  =  <^  =  a~^  =  a^  *,  by  §§  176,  115. 

(2)  (a*)"'  ..  -1-  =  JL  ^  a"^'  *,  by  §  177  and  (1). 

Hence  (a*^)"  =  a*^  for  all  rational  values  of  n  and  k. 


FRACTIONAL  AND  NEGATIVE  EXPONENTS     427 

To  verify  Law  IV.  Let  m  and  n  be  positive  or  negative  integers  or 
fractions,  and  let  r  and  s  be  positive  integers,  then  we  have 

r 

(1)  (a'^b'^y  =  </(^^¥y  =  v^a'*'"6«'- ,  by  §§  176,  115, 

=  </^  .  ^C^  =  J"  .  ft*"",  by  §§  120, 176. 

(2)  (a'»6")~'  =  — i- _  =         ^  =  a"* '"*  •  &"•  *",  by  §  177  and  (1). 

Hence  (a'^h'^'Y  =  aP'^hP^  for  all  rational  values  of  m,  n,  and  p. 

To  verify  Law  V.  We  have  (  — J  =  - —  for  all  rational  values  of 
m,  n,  and  p,  since 

I^Y  =  (a« .  &-»)p  =  a'^p  .  &-"P  =  — ,  §  177  and  Law  IV. 

180.  From  Laws  III,  IV,  and  V,  it  follows  that  any  mono- 
mial is  affected  with  any  exponent  by  multiplying  the  exponent 
of  each  factor  of  the  monomial  by  the  given  exponent. 


Ex.  1.    (ah-^c^y^  =  a~^  '  h~^ '   ^c  ^'^  =  a  h^c'^ 


Ex  2     /3flW"^^3"^a-ia:-8^   6^ 
^'    '    [   by^  J 

Ex  3     (l^yK(?IjlY^?llP!  =  ^ 


8i 
EXERCISES 

Perform  the  operations  indicated  by  the  exponents  in  each, 
of  the  following,  writing  the  results  without  negative  expo- 
nents and  in  as  simple  form  as  possible : 

1-  (M)"^-  5.  (x-^/yK  9.  (Y)'^.  13.  (.0009)1 

2.  (H)'^.  6.   25K  10.  (^)l  14.  (.027)i 

3.  (ff)'-  7.  25-\  11.  (0.25)i  15.  (32  a-'b^'^i 

4.  (27a-»)i  8.250.  12.  (0.25)-l  16.  8^  •  4"^ 


428  EXPONENTS  AND  RADICALS 

18.   (27!C«r''0"*-     20.   -V^fif -(M)"^-       ■   Va!-^2/A'»2/-V 


23.  -t'8rF^»(-27a%-'')-i      26.  V16a-6-«  •  VSa'ft"". 

24.  fJ!^y*  +  f?E_T*.    .27.  (-2-a-&-rk-2-Vi6-7 

29.  Prove  Law  III  in  detail  for  the  following  cases : 

(1)  (aY%    (2)  (a"")-,    (3)  (a'V'. 

30.  Prove  Law  IV  in  detail  for  the  following  cases : 

(1)  (a-*6-0%    (2)  (a-*6-0-i. 
Multiply : 

31.  x-^-i-x-^y-^  +  y-^  by  x-^-y-\ 

32.  x^  —  x^y^-\-y^  by  a;^  +  2/i 

33.  a?*  +  x^y^  +  aJ^^  +  x^y^  +  2/^  by  a?^  -  y^. 

34.  -x/^  +  a/^  by  </^--s/R 

35.  aj^  +  a;y  +  2/^  by  o;^  — i/^. 

36.  x  —  Sx^y~^  +  Sx^y-^-y~^  by  a?^  -  2  a; V^  +  2/"^. 

37.  x^  +  a?!/^  +  a'^2/^  +  2/^  ^7  ^^  —  2/^- 

Divide : 

38.  x^-x^^y  +  x^y-x^y^  +  x^y^-y*  by  x^-xhj-^-y. 

39.  3a^-a6*  +  4a62_3a^6  +  6^-4&3  by  3a^-6^  +  46='. 

40.  a^~3a;^+6aj^-7a;  +  6a;^-3a;*  +  l  by  a^^-a^^  +  l. 

41.  4a;V2-17a;*6''*  +  16a;-*6«  by  2  x^ -- b' -  4:  x-h\ 


REDUCTION  OF   RADICAL   EXPRESSIONS  429 

Find  the  square  root  of : 

42.  4  ar^  —  4  xy^^  +  4  xz~^  -\-y^  —  2  yh~^  +  z'\ 

43.  a-^-2a"M  +  6^  +  2a~V  +  c*-26V. 

44.  b-^-2  6-M  +  c*  +  2  6"id^  +  2  6"V^  -  2  cM  +  cfi 

4-2A"i-2cV^4-e-\ 
Find  the  cube  root  of :      45.  |  ^3  _  |  a^d^  +  6  a6  -  8  6l 

46.  a«-3a«  +  5a3-3a-l.  47.  a"^  +  3 a"*6^  +  3 a"*6^  +  6^. 

REDUCTION  OF  RADICAL  EXPRESSIONS 

181.  An  expression  containing  a  root  indicated  by  the  radical 
sign  or  by  a  fractional  exponent  is  called  a  radical  expression. 
The  expression  whose  root  is  indicated  is  the  radicand. 

E.g.  -y/E  and  (1  4-  xy  are  radical  expressions.  In  each  case  the 
index  of  the  radical  is  3. 

The  reduction  of  a  radical  expression  consists  in  changing 
its  form  without  changing  its  value. 

Each  reduction  is  based  upon  one  or  more  of  the  Laws  I  to 
V,  §  175,  as  extended  in  §  179. 

182.  To  remove  a  factor  from  the  radicand.  This  reduction  is 
possible  only  when  the  radicand  contains  a  factor  which  is  a 
perfect  power  of  the  degree  indicated  by  the  index  of  the  root, 
as  shown  in  the  following  examples : 

Ex.  1.    V72  =  \/36T2  =  VmV2  =  6  \/2. 

Ex.  2.  (a^xY)^  =  (oy  •  ^2)^  =  (aYy  •  i^^)^  =  ay^x^- 

This  reduction  involves  Law  IV,  and  may  be  written  in 
symbols  thus :  ,. 


430  EXPONENTS  AND   RADICALS 

EXERCISES 

In  the  expressions  on  p.  436,  remove  factors  from  the  radi- 
cands  where  possible. 

In  the  case  of  negative  fractional  exponents,  first  reduce  to  equivar 
lent  expressions  containing  only  positive  exponents. 

183.  To  introduce  a  factor  into  the  radicand.  This  process 
simply  retraces  the  steps  of  the  foregoing  reduction,  and  hence 
also  involves  Law  IV. 

Ex.  1.   6  V2  =  V62 .  V2  =  V36T2  =  \/72.  See  §  112 

Ex.  2.   afx^  =  V(^» .  v/^  =  v'(ay2)8a;2.=  V^^, 
Exi  3.   X  y/y  =  Vx"  Vy  =  Vx^. 

EXERCISES 

In  the  expressions  on  p.  436,  introduce  into  the  radicand  any 
factor  which  appears  as  a  coefficient  of  a  radical. 

184.  To  reduce  a  fractional  radicand  to  the  integral  form. 
This  reduction  involves  Law  IV  or  Law  V,  and  may  always  be 
accomplished. 

Ex.  1.    \/|=   Vif  =  V^Ti5=  ^vT5.  Law  IV 


Ex.2. 


Ex.  3.    —-  =  ^^—=^Q^2. 
vo  ^ 

T  ,    T  ,  rja       rjab''-^      -\/fl6''-^      1  ^, 

In  symbols,  we  have  \/ -  =  \  — r—  =  -   ,.—     =  t  ■N/a6''- 1. 

EXERCISES 

In  the  expressions  on  p.  436,  reduce  each  fractional  radicand 
to  the  integral  form. 

In  case  negative  exponents  are  involved,  first  reduce  to  equivalent 
expressions  containing  only  positive  exponents. 


REDUCTION  OF   RADICALS  431 

185.   To  reduce  a  radical  to  an  equivalent  radical  of  lower  index. 

This  reduction  is  effective  when  the  radicand  is  a  perfect  power 
corresponding  to  some  factor  of  the  index, 

Ex.1.    A/8  =  8^  =  8(a^2)  =  (8^)^  =  2i  =  V2. 


Ex.2,    ^a^  +  2 ab  +  b^  =  yj^a- -^2 cU) -{-b^  =  VcTfb. 

This  reduction  involves  Law  III  as  follows : 
11         11         1 

from  which  we  have 

''■Y]^=</i/^=VVi,   '  See  §114 

By  this  reduction  a  root  whose  index  is  a  composite  number 
is  made  to  depend  upon  roots  of  lower  degree. 

E.g.  A  fourth  root  may  be  found  by  taking  the  square  root  twice ; 
a  sixth  root,  by  taking  a  square  root  and  then  a  cube  root,  etc.  In 
the  case  of  literal  expressions  this  can  be  done  only  when  the  radicand 
is  a  perfect  power  of  the  degree  indicated  by  the  index  of  the  root. 

But  when  the  radicand  is  expressed  in  Arabic  figures,  such  roots  may 
in  any  case  be  approximated  as  in  §  127. 


EXERCISES 

In  the  expressions  on  p.  436,  make  the  reduction  above  indi- 
cated where  possible. 

In  the  case  of  arithmetic  radicands,  approximate  to  two  places  of  deci- 
mals such  roots  as  can  be  made  to  depend  upon  square  and  cube  roots. 

186.  To  reduce  a  radical  to  an  equivalent  radical  of  higher  index. 
This  reduction  is  possible  whenever  the  required  index  is  a 
multiple  of  the  given  index.     It  is  based  on  Law  III  as  follows : 

L  r    f  rt 

x^  =  {xsyt  =  x^K  See  §  179 

Ex.1.    Va=a^  =  (a^)^  =  a^  =  \/^. 
Ex.  2.    ^  =  fti  =  6*  =  -Vb^' 


432  EXPONENTS   AND   RADICALS 

Definition.  Two  radical  expressions  are  said  to  be  of  the 
same  order  when  their  indicated  roots  have  the  same  index. 

By  the  above  reduction  two  radicals  of  different  orders  may 
be  changed  to  equivalent  radicals  of  the  same  order,  namely,  a 
common  multiple  of  the  given  indices. 

E.g.    Va  and  v^  in  Exs.  1  and  2  above. 

EXERCISES 

In  Exs.  3,  4,  6,  17,  18,  23,  28,  30,  on  p.  436,  reduce  the 
corresponding  expressions  in  the  first  and  second  columns  to 
equivalent  radicals  of  the  same  order. 

187.  In  general,  radical  expressions  should  be  at  once 
reduced  so  that  the  order  is  as  low  as  possible  and  the  radicand 
is  integral  and  as  small  as  possible.  A  radical  is  then  said  to 
be  in  its  simplest  form. 

ADDITION  AND  SUBTRACTION  OF  RADICALS 

188.  Definition.  Two  or  more  radical  expressions  are  said 
to  be  similar  when  they  are  of  the  same  order  and  have  the 
same  radicands. 

E.g.  3V7  and  5V7   are   similar  radicals  as   are   also  aVx^  and 

If  two  radicals  can  be  reduced  to  similar  radicals,  they  may 
be  added  or  subtracted  according  to  §  10. 

Ex.  1.     Find  the  sum  of  V8,  V50,  and  V98. 

By  §  182,  V8  =  2  V2,  V50  =  5  V2,  and  \/98  =  7  V2. 

Hence  \/8+V50+V98  =  2\/2  +  5V2  +  7\/2  =  UV2. 

Ex.  2.    Simplify  VJ  -  V20  +  V3|. 

By  §184,  Vl=^VE,  v^=2V5,V3l=v^  =  4\/|  =  |\/5. 

Hence  Vi-V20+V3i=W5-2\/6  +  |>/5  =  -\/5. 


MULTIPLICATION  OF  RADICALS  483 

If  two  radicals  cannot  be  reduced  to  equivalent  similar  radi- 
cals, their  sum  can  only  be  indicated. 

E.g,    The  sum  of  V2  and  \/5  is  V2  +  v^B. 

Observe,  however,  that 

VIO  +  \/6  =  v^.>/5  +  \/2.V3=V2(V5+  \/3). 

EXERCISSS 

(a)  In  Exs.  1,  2,  5,  7,  8,  19,  20,  21,  p.  436,  reduce  each 
pair  so  as  to  involve  similar  radicals  and  add  them. 

(6)   Perform  the  following  indicated  operations : 

1.  V28  +  3V7-2V63.  5.    V|+V63  +  5V7. 

2.  \/24--v/8T-^^.  6.   V99-11VS+V44. 

3.  a/^+ ^/^^  - -\/32^.  7.   2V?  +  3V|^4-Vi76. 

4.  2V48-3V12  +  3V4.  8.    V^  +  6VJ-Vl2. 

9.    -^9 +  ^27  4- -^5^^=^. 
10.  {x"  4- 1)  Va^  H-  arb  -  V(a2  -.}^){a-h), 

MULTIPLICATION   OF  RADICALS 

189.  Radicals  of  the  same  order  are  multiplied  according  to 
§  120  by  multiplying  the  radicands.  If  they  are  not  of  the  same 
order,  they  may  be  reduced  to  the  same  order  according  to  §  186. 

E.g.     yTa-y/h^  ah^  =  aM  =  v/^8  .  v^  =  v^sp. 
In  many  cases  this  reduction  is  not  desirable.     Thus,  Vx^ .  V^  is 
written  x^y^  rather  than  y/x^y^. 

Radicals  are  multiplied  by  adding  exponents  when  they  are 
reduced  to  the  same  base  with  fractional  exponents,  §  176. 

E.g.     ^/F^.V^  =  x^.x^  =  x^'^^  =  xJ^. 


434  EXPONENTS   AND   RADICALS 

190.   The  principles  just  enumerated  are  used  in  connection 
with  §  10  in  multiplying  polynomials  containing  radicals. 

Ex.1.   3  a/2  4- 2  V5  Ex.  2.   3  V2  +  2  V5 

2V2-3V5  3V2-2V5 


6-2    4-4V10  .  9-2    +6V10 

-  9  VlO  -6.5  -  6  Vip  -4.5 

12      -5Vi0-30  18  -20 

Hence  (2  V2  +  2  V5)(2  V2  -  3  V5)  =  -  18  -  5  VlO, 

and  (3  V2  +  2  V5)(3  ■^J2-2  V5)  =  18  -  20  =  -2. 

EXERCISES 

(a)    In  Exs.  21  to  38,  p.  436,  find  the  products   of  the 
corresponding  expressions  in  the  two  columns. 

(6)   Find  the  following  products: 

1.  (3+>Vri)(3- Vii). 

2.  (3  V2  +  4  V5)(4  V2  -  5  V5). 

3.  (2  +  V3  +  V5)(3  +  V3  -  V5). 

4.  (3  V2  -  2  Vi8  +  2  ■\/'l){^^2-  Vl8  -  V7). 

5.  (Va  -  V6)(V^  +  V6)(a2  +  a&  +  &')• 

6.  (Vvi3  +  3)(VVI3  -  3). 

7.  (V2  +  3  V5)(V2  +  3  V5). 

8.  (3a-2V^)(4a  +  3  V^). 

9.  (3  V3  +  2  V6  -  4  V8)(3  V3  -  2  V6  +  4  V8). 
10.  ( Va  +  V&  -  ■\/~c){^-\/~(i  -  V6  +  Vc). 


DIVISION  OF  RADICALS  435 

11.  (a-Vb-  ^c){a  +  V6  +  Vc). 

12.  (2V|  +  3V|4-4V|)(2V|-5Vf). 

13.  {Va'  +  W){V^^  +  ^'^^  +  -W)' 

14.  (</^-2/')3. 

DIVISION  OF  RADICALS 

191.  Radicals  are  divided  in  accordance  with  Laws  II  and  V. 
That  is,  the  exponents  are  subtracted  when  the  bases  are  the 
same,  and  the  bases  are  divided  when  the  expoiients  are  the  same. 
See  §§  179, 121. 

Ex.1.    ^^-V^  =  a;?-^a;^  =  ic^"^=a;■**. 
Ex.  2.   x^  -  y^  =  ^-V  =  (a;rO^=  \/^p. 

Ex.3.   V^-^^6  =  a^^6^  =  (^^y=^^^6^. 

EXERCISES 

(a)  In  each  of  the  Exs.  1  to  20,  on  p.  436,  divide  the 
expression  in  the  first  column  by  that  in  the  second. 

(6)   Perform  the  following  divisions : 

1.  (Vo^  +  2  Va^  -  3  Va)  --  6  Va. 

2.  ( Va  +  -v/6  -  c)  -^  Vc. 

3.  (2  VO  4- 3  Vl2  -  4  Vl5) -s- VB. 

4.  (4V7-8V2i+6V42)--2V7. 


436 


EXPONENTS  AND   RADICALS 
EXERCISES 


1. 

3V45, 

2V125. 

21. 

3(50)*, 

4(72)i. 

2. 

ai 

ai 

22. 

4 

at. 

3. 

-i 

a=l 

23. 

ax^y 

daji 

4. 

3Va^t/, 

2V^. 

24. 

ah\ 

al6^. 

5. 

d^/a'b% 

c</^\ 

25. 
26. 

ma»% 

6. 

TV  (a -{-by, 

llV(a-hf. 

S-t^aWc". 

7. 

</a^ 

ai. 

27. 

Va'W, 

-?/6»</S^. 

8. 

wVm*, 

ml. 

28. 

8t, 

16*. 

9. 

Vf. 

V?. 

29. 

25-^ 

125-t. 

10. 

vh, 

V|f. 

30. 

9^, 

8i 

11. 
12. 

vp, 

1 

a-V, 

31. 
32. 

13. 

33. 

3a-W 

4a-VI 

14. 


3a-^ 


6* 

4 
2a*5-2 


xo. 

c-%^' 

c^ 

16. 

n 

n 

17. 

n 

n 

18. 

■u, 

n 

19. 

18*, 

V32. 

20. 

Vl2, 

48i 

34. 


c-id~^  dV 


35.  5(a+6)-t,    3(a4-6)'i 

36.  (-64)^,        -64i 

37.  (^)*,        (^yK 


</¥' 


38.  i> 

39.  ■\/^a^-12a'b-\-12ab''-4:b\ 

40.  V(3a-26)(9a2-4  62). 


RATIONALIZING  THE   DIVISOR  437 

192.  Rationalizing  the  divisor.  In  case  division  by  a  radical 
expression  cannot  be  carried  out  as  in  the  foregoing  examples, 
it  is  desirable  to  rationalize  the  denominator  when  possible. 

Ex.1.    V^^V2-V5^V10. 
V5      V5-V5        5 

Ex  2         V«       _        Va(Va  +  V^)        _a+V«&, 
V^_V6     (Va-V6)(Va-|rV6)        «-^ 

Evidently  this  is  always  possible  when  the  divisor  is  a  mono- 
mial or  binomial  radical  expression  of  the  second  order. 

The  number  by  which  numerator  and  denominator  are  multi- 
plied is  called  the  rationalizing  factor. 

For  a  monomial  divisor,  Vx,  it  is  Vx  itself.  For  abinoriiial  divisor, 
Vxi  Vy,  it  is  the  same  binomial  with  the  opposite  sign,  y/xT  Vy. 


EXERCISES 

Eeduce  each  of  the  following  to  equivalent  fractions  having 
a  rational  denominator. 


1. 


2.      ^_^     ^.  ^    3V3--2V2 

-^'^  '  -^  '   3V3+2V2 


2- V5' 

7 

V5  +  V3* 

V27 

V3  +  Vii 

2-V7 

2  +  V7 

V2-V3 

Vg  +  Vy. 

Va;  —  Vy  * 


8. 


4.    =^ ^~  9. 


Va^  -  1  +  Va*  +  1 


Va;  +  1  +  V 


iC- 


Va;  +  1  -  Va;  - 1 

g      ,  -  ^^     Va  —  6  —  Va  -h  &. 

*    V2  +  V3  '    Vo^^  +  VoTd 


438.  EXPONENTS   AND    RADICALS 

193.  In  finding  the  value  of  such  an  expression  as 
— — — — ^,  the  approximation  of  two  square  roots  and  division 

by  a  decimal  fraction  would  be  required.     But  "^ '  +  "^f  equals 

114-2V2I       .  .  V7-V3 

\   - — — which  requires  only  one  root  and  division  by  the 

4 

integer  4. 

EXERCISES 

Find  the  approximate  values  of  the  following  expressions  to 
three  places  of  decimals. 

^    3V5  +  4V3  g    7V5  +  3V8^ 

V5-V3  '   2V5-3V2* 

V7  g    5\/l9-3V7 

3V7-V19 

^    3V2- V5 
V5-6V2 

^    ^^  .  ^     ^  V  ^  g    5V6-7Vi3 

2V5+V3  ■    3V13-7V6' 

194.  Square  root  of  a  radical  expression.  A  radical  expression 
of  the  second  order  is  sometimes  a  perfect  square,  and  its 
square  root  may  be  written  by  inspection. 

E.g.  The  square  of  Va  ±  Vb  is  a  +  6  ±  2  Vab.  Hence  if  a  radical 
expression  can  be  put  into  the  form  x  ±  2  V/,  where  x  is  the  sum  of 
two  numbers  a  and  b  whose  product  is  y,  then  Va  ±  Vb  is  the  square 
root  of  a:  +  2  Vy. 

Example.     Find  the  square  root  of  5  4-  V24. 

Since  5  +  V24  =  5  +  2  V6,  in  which  5  is  the  sum  of  2  and  3,  and  6  is 
their  product,  we  have  V5  +  ^24  =  V2  +  VS. 


V7-V2 

4V3 

V3  -  V2 

11V5-3V3 

IMAGINARIES 


439 


Find  the  square  root  of  each  of  the  following : 

1.  3-2V2.  3.   8-V60.  5.  24-6V7. 

2.  7  +  ViO.  4.   7  4-  4  V3.  6.   28  +  3  vl2. 

195.  Radical  expressions  involving  imaginaries.  According  to 
the  definition,  §  112,  (V^^)^  =  -  1.  Hence,  (V^^^)^ 
=(V^:n:)V^i  =  -V^  and  (V^I)^  =  (V^XV^^)' 
=  (-l)(-l)=  +  l. 

The  following  examples  illustrate  operations  with  radical 
expressions  containing  imaginaries. 


Ex.  1. 


4  + V-9 


V4V-1+V9V^^ 
=  (2  4-  3)  V^^  =  5  V^l. 

Ex.2.    V"=^- V^^  =  V4- V9-(V^2^-2.3  =  -6. 
4      V4V~ 


Ex.3. 


V 


V4^2 
V9     3* 


V-9      V9V-1 

Ex.  4.    V^r2  .  V^^  •  V^6  =  V2  .  V3  •  V6  .  (V^^)^ 

=  _V36V^^=-6V^. 
Ex.  5.    Simplify  (J  +  ^ V^)^ 


We   are  to  use   ^(l+v^V-l)  three  times  as  a  factor, 
serving  (i)^  =  ^  as  the  final  coefficient,  we  have, 


Re- 


1  + V3V- 
l+ V3V- 

l  +  V3  V  - 

V3V- 

T-3 

-2  +  2\/3V- 

1+      y/Sy^ 

-2  +  2V3  V- 
-2v^\/- 

3-6 

Hence  Q  +  ^  V^^)8  =  K-  8) 


-6 


8. 


440  .  EXPONENTS   AND   RADICALS 

EXERCISES 

Perform  the  following  indicated  operations. 


1. 

V- 

-16-h  V- 

-9  +  V^ 

-25. 

2. 

v^ 

.x'--V~^ 

-x". 

3. 

3  +  5V-1- 

-2V-1. 

4.   (2  +  3V-l)(3  +  2V-l). 


5.  (2  +  3V-l)(2-3V-l). 

6.  (4-f  5V^^)(4-5V^. 

7.  (2V2-3  v/^=^)(3V3H-2V^=^). 

8.  (V^^4-V^=^)(V^3-V^^). 

9.  (3V5-f-2V^:7)(2V5-3  V^. 
10.  (-i_iV^)(-i-iV^)^. 

Eationalize  the  denominators  of 


11. 


12. 


13. 


1- 

-V- 
3 

-1 

V3+^ 

/-I 

1- 

-V^ 

T 

14. 


15. 


16. 


V2+V-3 


V2-V- 
6 

-3 

2-3V- 

■5 

1+V-i 


;V-l-2/ 


17.  Solve  a;*  —  1  =  0  by  factoring.    Eind  four  roots  and  verify 
each. 

18.  Solve  ar^  4- 1  =  0  by  factoring  and  the  quadratic  formula. 
Find  three  roots  and  verify  each. 

19.  Solve  ic^  —  1  =  0  as  in  the  preceding  and  verify  each  root. 

20.  Solve  ic^  ~  1  ==  0  by  factoring  and  the  quadratic  formula. 


EQUATIONS   CONTAINING   RADICALS  441 

SOLUTION  OF  EQUATIONS   CONTAINING  RADICALS 

196.  Many  equations  containing  radicals  are  reducible  to 
equivalent  rational  equations  of  the  first  or  second  degree. 

The  method  of  solving  such  equations  is  shown  in. the  fol- 
lowing examples. 

Ex.  1.   Solve  1  +  Vx  =  V3  +  X.  (1) 

Squaring  and  transposing,       2V'a;  ^  2.  (2) 

Dividing  by  2  and  squaring,  x  =  1.  (3) 

Substituting  in  (1),  1  +  1  =  VsTl  =  2. 

Observe  that  only  principal  roots  are  used  in  this  example. 

If  (1)  is  written  1  +  Vi  =  -  V3  +  a;,  (4) 

then  (2)  and  (3)  follow  as  before,  but  x  =  1  does  not  satisfy  (4).  In- 
deed equation  (4)  has  no  root.  This  should  not  be  confused  with  the 
fact  that  every  integral,  rational  equation  has  at  least  one  root. 


Ex.  2.   Solve  Va;  +  5  =  a;  - 1.  (1) 

Squaring  and  transposing,        a;^  —  3  a;  —  4  =  0.  (2) 

Solving,  a;  =  4  and  a;  =  - 1. 

a;  =  4  satisfies  (1)  if  the  principal  root  in  Va:  -f  5  is  taken.  a:=  —1 
does  not  satisfy  (1)  as  it  stands  but  would  if  the  negative  root  were 
taken. 


Ex.3.   Solve      V4^+2-V3^^^1  ^^^ 

V4a;  +  l  +  V3a;-2     ^ 
Clearing  of  fractions  and  combining  similar  radicals. 


2V4ar-f-l  =  3V3a:-2.  (2) 

Squaring  and  solving,  we  find  a:  =  2. 

This  value  of  x  satisfies  (1)  when  all  the  roots  are  taken  positive  and 
also  when  all  are  taken  negative,  but  otherwise  not. 


442  ^        EXPONENTS   AND   RADICALS 

Ex.  4.   Solve  V2  aj  +  3  =  — =:- — 7  -  1-  (1) 

V3a;— 1 

The  fraction  in  the  second  member  should  be  reduced  as  follows: 

— ;= =  ^ -;=± '   =  V3  a;  +  1. 

V3a:-1  V3x-1 

Hence,  (1)  reduces  to     \/2  x  +  3  =  V3^  +  l  _  1  ^  V3I.  (2) 

Solving,  a:  =  3,  which  satisfies  (1). 

If  we  clear  (1)  of  fractions  in  the  ordinary  manner,  we  have 


( V3  x-l)  V2x  +  3  =  -  V3  a;  +  3  a:.  (2') 

Squaring  both    sides   and  transposing   all  rational   terms  to  the 

second  member,  

2  a;  \/3  a;  -  6  \/3^  =  3  a:^  -  8  a;  -  3.  (3) 

Factoring  each  member, 

2(a;  -  3)  V3^  ^  {x  -  3)(3  x  +  1),  (4) 

which  is  satisfied  by  a;  =  3. 

Dividing  each  member  by  a;  —  3,  squaring  and  transposing,  we  have 
9  a;2  -  6  a;  +  1  =  (3  a;  -  1)2  =  0,  (5) 

which  is  satisfied  by  a;  =  |. 

Equation  (1)  is  not  satisfied  hy  x=\,  since  the  fraction  in  the 
second  member  is  reduced  to  ^  by  this  substitution.  See  §  50. 
The  root'  x  =  \  is  introduced  by  clearing  of  fractions  without 
first  reduci7ig  the  fraction  to  its  loivest  terms,  V3  x  —  1  being  a 
factor  of  both  the  numerator  and  the  denominator.     See  §  165. 

Ex.5.     Solve  -«^^-V3  =  ^^.  (1) 

V6  —  X  Va?  —  3 

Reducing  the  fractions  by  removing  common  factors,  we  have 

VqZTx  -V3=  V^^^.  (2) 

Squaring,  transposing,  and  squaring  again, 

a;2  _  9  a:  +  18  =  0,  (3) 

whence  a:  =  3,   ar  =  6. 

But  neither  of  these  is  a  root  of  (1).     In  this  case  (1)  has  no  root. 


EQUATIONS   CONTAINING   RADICALS  443 

197.  In  solving  an  equation  containing  radicals,  we  note  the 
following : 

(1)  If  a  fraction  of  the  form       ^~ — -  is  involved,  this 

Va  — Vft 
should  be  reduced  by  dividing  numerator  and  denominator  by 

Va  —  Vft  before  clearing  of  fractions. 

(2)  After  clearing  of  fractions,  transpose  terms  so  as  to 
leave  one  radical  alone  in  one  member. 

(3)  Square  both  members,  and  if  the  resulting  equation  still 
contains  radicals,  transpose  and  square  as  before. 

(4)  In  every  case  verify  all  results  by  substituting  in  thft 
given  equation.  In  case  any  value  does  not  satisfy  the  given 
equation,  determine  whether  the  roots  could  be  so  taken  that 
it  would.     See  Ex.  3. 

EXERCISES 

Solve  the  following  equations  : 


1.    Va;2-|-7a;-2-Va^-3a;  +  6  =  2. 

5 


2.    V3?/-V32/-7.= 


VSy-7 


V62/  +  1  2  *  V5a;  +  1 


X  X 


4.    V5aj-19  +  V3a;  +  4  =  9.      ^  4      VT^T^        ^ 

^'  Z 1 ~  V 3. 

Vfl^  +  CT^  —  a;  _  2 
V^+^  +  a; 


6.  ^a-{-Va^+^=Va—Vx.  4  +  a;  — VSaj  +  ic^ 

7.  J=lL_=Vi^+2VZ.   11.     ^-^  ^  ^+«  =V^^36: 
V2/H-V^  ^  -\/a  —  x     Vaj+a 

12.       ^-g  ^V^  +  V^  +  2Va. 
Va;  —  Va  ^ 


444  EXPONENTS   AND   RADICALS 

13.  —      ^    —wm  —  n  = ^ 

Vm  —  y  -y/y  —  n 

14.  2V^"='^  +  3V2^  =  '^^±^. 

Va;  — a 

15.  V2a;  +  7+V2a;  +  14=V4a;  +  35  +  2V4ic2  +  42a;^^2i. 


16.  Vx  —  S  +  -Vx-{-9==Vx  +  TS  +  Vx—e. 

17.  VaJ  +  T-Va;  — l=Va  +  2+Va;  — 2. 


18.   aV2/  +  6- cV6-2/  =  V6(a^4-c^. 


19.   2/V2/— c  — V2/^  +  c*  +  cV^H-c  =  0.  ^ 
Vm      Va;      V^      Vm 


21.  ^|u+Vx+^i6-'Vx= 


12 


22.    V3aJ4-V3a;  +  13 


91 


V3.r  +  13' 


23.  V6a;4-3  + Va;  +  3  =  2a;+3. 

24.  -y/x  —  a  +  V&  — a;=  -Vb  —  a. 
■\/x  —  a  +  Vic—  ft 


a;  — a  + Va;—  ft  _    jx—a 
x  —  a  —  -Vx  —  ft      ^aj  — ft* 

26.    V(a;-l)(aj-2)  +  V(a;— 3)(a;  -4)  =  V2. 


25. 

V 


27.    V2  a;  +  2  +  V7  +.6  aj  =  V7  a;  +  72. 


28.    V2  afta;  +  Va^  —  fta;  =  Va^  +  6a;. 

a  4-  a;  +  Va^  —  a^     c 


29. 


a  +  aj-Va^-ar'     x 
30.    Vaj2-2a;  +  4+V3a^  +  6a;  +  12  =  2Va^  +  a;  +  10. 


PROBLEMS 


445 


PROBLEMS 

1.  Find  the  altitude  drawn  to  the  longest  side  of  the  tri- 
angle whose  sides  are  6,  7,  8. 

Hint.  See  figure,  p.  190,  E.  C.  Calling  x  and  S  —  x  the  segments 
of  the  base  and  h  the  altitude,  set  up  and  solve  two  equations  involv- 
ing X  and  h. 

2.  Find  the  area  of  a  triangle  whose  sides  are  15,  17,  20. 
First  find  one  altitude  as  in  problem  1. 

3.  Find  the  area  of  a  triangle  whose  base  is  16  and  whose 
sides  are  10  and  14. 

4.  Find  the  altitude  on  a 
side  a  of  a  triangle  two  of 
whose  sides  are  a  and  a  third  b. 

A  three-sided  pyramid  all  of 
whose  edges  are  equal  is  called  a 
regular  tetrahedron.  In  Figure  10 
AB,AC,AD,BC,  BD,  CD  are  all 
equal. 

5.  Find  the  altitude  of  a 
regular  tetrahedron  whose  edges 
are  each  6.  Also  the  area  of  the 
base. 

Hint.  First  find  the  altitudes  AE  and  DE  and  then  find  the  alti- 
tude of  the  triangle  AED  on  the  side  DE,  i.e.  find  AF. 

6.  Find  the  volume  of  a  regular  tetrahedron  whose  edges 
are  each  10. 

The  volume  of  a  tetrahedron  is  ^  the  product  of  the  base  and 
the  altitude. 

7.  Find  the  volume  of  a  regular  tetrahedron  whose  edges 
are  each  a. 

8.  In  Figure  10  find  EG  if  the  edges  are  each  a. 

9.  If  in  Figure  10  EG  is  12,  compute  the  volume. 

Use  problem  8  to  find  the  edge,  then  use  problem  7  to  find  the 
volume. 


Fig.  10. 


446  EXPONENTS   AND   RADICALS 

10.  Express  the  volume  of  the  tetrahedron  in  terms  of  EG, 
That  is  if  EG=h,  find  a  general  expression  for  the  volume  in 
terms  of  h. 

11.  If  the  altitude  of  a  regular  tetrahedron  is  10,  compute 
the  edge  accurately  to  two  places  of  decimals. 

12.  Express  the  edge  of  a  regular  tetrahedron  in  terms  of 
its  altitude. 

13.  Express  the  volume  of  a  regular  tetrahedron  in  terms 
of  its  altitude. 

14.  Express  the  edge  of  a  regular  tetrahedron  in  terms  of 
its  volume. 

15.  Express  the  altitude  of  a  regular  tetrahedron  in  terms 
of  its  volume. 

16.  Express  EG  of  Figure  10  in  terms  of  the  volume  of  the 
tetrahedron. 

17.  Eind  the  edge  of  a  regular  tetrahedron  such  that  its  vol- 
ume multiplied  by  V2,  plus  its  entire  surface  multiplied  by  V3, 
is  144. 

The  resulting  equation  is  of  the  third  degree.     Solve  by  factoring. 

18.  An  electric  light  of  32  candle  power  is  25  feet  from  a 
lamp  of  6  candle  power.  Where  should  a  card  be  placed 
between  them  so  as  to  receive  the  same  amount  of  light  from 
each? 

Compare  problem  13,  p.  423.  Compute  result  accurately  to  two 
places  of  decimals. 

19.  "Where  must  the  card  be  placed  in  problem  18  if  the 
lamp  is  between  the  card  and  the  electric  light? 

Notice  that  the  roots  of  the  equations  in  18  and  19  are  the  same. 
Explain  what  this  means. 

20.  State  and  solve  a  general  problem  of  which  18  and  19 
are  special  cases. 


PROBLEMS  447 

21.  If  the  distance  between  the  earth  and  the  sun  is  93 
million  miles,  and  if  the  mass  of  the  sun  is  300,000  times  that 
of  the  earth,  find  two  positions  in  which  a  particle  would  be 
equally  attracted  by  the  earth  and  the  sun. 

The  gravitational  attraction  of  one  body  upon  another  varies  in- 
versely as  the  square  of  the  distance  and  directly  as  the  product  of 
the  masses.     Represent  the  mass  of  the  earth  by  unity. 

22.  Find  the  volume  of  a  pyramid  whose  altitude  is  7  and 
whose  base  is  a  regular  hexagon  whose  sides  are  7. 

The  volume  of  a  pyramid  or  a  cone  is  ^  the  product  of  its  base  and 
its  altitude. 

23.  If  the  volume  of  the  pyramid  in  problem  22  were  100 
cubic  inches,  what  would  be  its  altitude,  a  side  of  the  base 
and  the  altitude  being  equal  ?  Approximate  the  result  to  two 
places  of  decimals. 

24.  Express  the  altitude  of  the  pyramid  in  problem  22  in 
terms  of  its  volume,  the  altitude  and  the  sides  of  the  base 
being  equal. 

25.  If  in  a  right  prism  the  altitude  is  equal  to  a  side  of  the 
base,  find  the  volume,  the  base  being  an  equilateral  triangle 
whose  sides  are  a. 

The  volume  of  a  right  prism  or  cylinder  equals  the  product  of 
its  base  and  its  altitude. 

26.  Find  the  volume  of  the  prism  in  problem  25  if  its  base 
is  a  regular  hexagon  whose  side  is  a. 

27.  Express  the  side  of  the  base  of  the  prism  in  problem  25 
in  terms  of  its  volume.  State  and  solve  a  particular  problem 
by  means  of  the  formula  thus  obtained. 

28.  Express  the  side  of  the  base  of  the  prism  in  problem  26 
in  terms  of  its  volume.  State  and  solve  a  particular  problem 
b^  means  of  the  formula  thus  obtained. 


448 


EXPONENTS  AND   RADICALS 


In  Figures  11  and  12  the  altitudes  are  each  supposed  to  be 
three  times  the  side  a  of  the  regular  hexagonal  bases. 


Fig.  11. 


Fig.  12. 


29.  Express  the  difference  between  the  volumes  of  the  pyra- 
mid and  the  circumscribed  cone  in  terms  of  a. 

The  volume  of  a  cone  equals  \  the  product  of  its  base  and  altitude. 

30.  Express  a  in  terms  of  the  difference  between  the  volumes 
of  the  cone  and  pyramid.  State  and  solve  a  particular  problem 
by  means  of  the  formula  thus  obtained. 

31.  Express  the  volume  of  the  pyramid  in  terms  of  the  dif- 
ference between  the  areas  of  the  bases  of  the  cone  and  the 
pyramid.  State  a  particular  case  and  solve  by  means  of  the 
formula  first  obtained. 

The  lateral  area  of  a  right  cylinder  or  prism  equals  the  perimeter 
of  the  base  multiplied  by  the  altitude. 

.  32.   Express  the  difference  of  the  lateral  areas  of  the  cylin- 
der and  the  prism  in  terms  of  a. 

The  following  four  problems  refer  to  Figure  12.  In  each  case  state 
a  particular  problem  and  solve  by  means  of  the  formula  obtained. 

33.  Express  a  in  terms  of  the  difference  of  the  lateral  areas. 

34.  Express  the  volume  of  the  prism  in  terms  of  the  differ- 
ence of  the  perimeters  of  the  bases. 

35.  Express  the  volume  of  the  cylinder  in  terms  of  the  dif- 
ference of  the  lateral  areas. 

36.  Express  the  sum  of  the  volumes  of  the  prism  and  cylin- 
der in  terms  of  the  difference  of  the  areas  of  the  bases. 


CHAPTER   XI 
LOGARITHMS 

198.  The  operations  of  multiplication,  division,  and  finding 
powers  and  roots  are  greatly  shortened  by  the  use  of  logarithms. 

The  logarithm  of  a  number,  in  the  system  commonly  used, 
is  the  index  of  that  power  of  10  which  equals  the  given  number. 

Thus,  2  is  the  logarithm  of  100  since  10^  =  100. 
This  is  written  log  100  =  2. 

Similarly  log  1000  =  3,  since  10«  =  1000, 

and  log  10000  =  4,  since  10*  =  10000. 

The  logarithm  of  a  number  which  is  not  an  exact  rational 
power  of  10  is  an  irrational  number  and  is  written  approxi- 
mately as  a  decimal  fraction. 

Thus,  log  74  =  1.8692  since  lOi-s^oa  =  74  approximately. 

In  higher  algebra  it  is  shown  that  the  laws  for  rational  ex- 
ponents (§  179)  hold  also  for  irrational  exponents. 

199.  The  decimal  part  of  a  logarithm  is  called  the  mantissa, 
and  the  integral  part  the  characteristic. 

Since  10«  =  1,  10^  =  10,  10^  =  100,  10^  =  1000,  etc.,  it  follows 
that  for  all  numbers  between  1  and  10  the  logarithm  lies  be- 
tween 0  and  1,  that  is,  the  characteristic  is  0.  Likewise  for 
numbers  between  10  and  100  the  characteristic  is  1,  for  num- 
bers between  100  and  1000  it  is  2,  etc. 

200.  Tables  of  logarithms  (see  p.  452)  usually  give  the  man-" 
tissas  only,  the  characteristics  being  supplied,  in  the  case  of 
whole  numbers,  according  to  §  199,  and  in  the  case  of  decimal 
numbers,,  as  shown  in  the  examples  given  under  §  201. 

449 


450  LOGARITHMS 

201.  An  important  property  of  logarithms  is  illustrated  by 
the  following: 

From  the  table  of  logarithms,  p.  470,  we  have : 

log  376  =  2.5752,  or  376  =  102-5752.  (1) 

Dividing  both  members  of  (1)  by  10  we  have 

37.6  =  102-5752  ^  101  ^  102.5752-1  ^  101.5752, 

Hence,  log  37.6  =  1.5752, 

Similarly,  log  3.76  =  1.5752  -  1  =  0..5752, 

log  .376  =  0.5752  -  1,  or  1.5752, 
log  .0376  =  0.5752  -  2,  or  2.5752, 

where  I  and  2  are  written  for  —  1  and  —  2  to  indicate  that  the  char- 
acteristics are  negative  while  the  mantissas  are  positive. 
Multiplying  (1)  by' 10  gives 

log  3760    =  2.5752  +  1  =  3.5752, 
and  log  37600  =  2.5752  +  2  =  4.5752. 

Hence,  if  the  decimal  point  of  a  number  is  moved  a  certain 
number  of  places  to  the  7'ight  or  to  the  left,  the  characteristic  of 
the  logarithm  is  increased  or  decreased  by  a  corresponding 
number  of  units,  the  mantissa  remaining  the  same. 

From  the  table  on  pp.  452,  453,  we  may  find  the  mantissas 
of  logarithms  for  all  integral  numbers  from  1  to  1000.  In  this 
table  the  logarithms  are  given  to  four  places  of  decimals, 
which  is  sufficiently  accurate  for  most  practical   purposes. 

E.g.  for  log  4  the  mantissa  is  the  same  as  that  for  log  40  or  for  log 
400. 

To  find  log  .0376  we  find  the  mantissa  corresponding  to  376,  and 
prefix  the  characteristic  2.     See  above. 

.     Ex.  1.   Find  log  876. 

Solution.  Look  down  the  column  headed  iV  to  87,  then  along  this 
line  to  the  column  headed  6,  where  we  find  the  number  9425,  which  is 
the  mantissa.      Hence  log  876  =  2.9425. 


LOGARITHMS  451 

Ex.  2.   Find  log  3747. 

Solution.   As  above  we  find  log  3740  =  3.5729, 
and  log3750  =  3.5740. 

The  difference  between  these  logarithms  is  11,  which  corresponds  to 
a  difference  of  10  between  the  numbers.  But  3740  and  3747  differ 
by  7.  Hence,  their  logarithms  should  differ  by  ^^  of  11,  i.e.  by  8.1. 
Adding  this  to  the  logarithm  of  3740,  we  have  3.5737,  which  is  the 
required  logarithm. 

The  assumption  here  made,  that  the  logarithm  varies  directly 
as  the  number,  is  not  quite,  but  v^y  nearly,  accurate,  when  the 
variation  of  the  number  is  confined  to  a  narrow  range,  as  is 
here  the  case. 

Ex.  3.    Find  the  number  whose  logarithm  is  2.3948. 

Solution.  Looking  in  the  table,  we  find  that  the  nearest  lower  loga- 
rithm is  2.3945  which  corresponds  to  the  number  248.     See  §  199. 

The  given  mantissa  is  3  greater  than  that  of  248,  while  the  man- 
tissa of  249  is  17  greater.  Hence  the  number  corresponding  to 
2.3948  must  be  248  plus  ^j  or  .176.  Hence,  248.18  is  the  required 
number,  correct  to  2  places  of  decimals. 

Ex.4.   Find  log  .043. 

Solution.   Find  log  43  and  subtract  3  from  the  characteristic. 

Ex.  5.   Find  the  number  whose  logarithm  is  4.3949. 

Solution.  Find  the  number  whose  logarithm  is  0.3949,  and  move 
the  decimal  point  4  places  to  the  left. 

EXERCISES 

Find  the  logarithms  of  the  following  numbers : 

1.  491.  6.   .541.  11.   .006.  16.  79.31. 

2.  73.5.  7.    .051.  12.   .1902.  17.  4.245. 

3.  2485.  8.   8104.  13.    .0104.  18.  .0006. 

4.  539.7.  9.   70349.    .  14.   2.176.  19.  3.817. 

5.  53.27.  10.   439.26.  15.   8.094.  20.  .1341. 


452 


LOGARITHMS 


N 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

10 

11 

12 
13 
14 

0000 
0414 
0792 
1139 
1461 

0043 
0453 
0828 
1173 
1492 

0086 
0492 
0864 
1206 
1523 

0128 
0531 
0899 
1239 
1553 

0170 
0569 
0934 
1271 
1584 

0212 
0607 
0969 
1303 
1614 

0263 
0645 
1004 
1336 
1644 

0294 
0682 
1038 
1367 
1673 

0334 
0719 
1072 
1399 
1703 

0374 
0756 
1106 
1430 
1732 

15 

16 
17 
18 
19 

1761 
2041 
2304 
2553 

2788 

1790 
2068 
2330 

2577 
2810 

1818 
2095 
2355 
2601 
2833 

1847 
2122 

2380 
2625 
2856 

1876 
2148 
2406 
2648 
2878 

1903 
2176 
2430 

2672 
2900 

1931 
2201 
2465 
2695 
2923 

1959 
2227 
2480 
2718 
2945 

1987 
2253 
2604 

2742 
2967 

2014 
2279 
2529 
2766 
2989 

20 

21 
22 
23 
24 

3010 
3222 
3424 
3617 
3802 

3032 
3243 
3444 
3636 
3820 

3054 
3263 
3464 
3655 
3838 

3075 
3284 
3483 
3674 
3856 

8096 
3304 
3502 
3692 
3874 

3118 
3324 
3622 
3711 

3892 

3139 
3345 
3641 
3729 
3909 

3160 
3365 
3660 
3747 
3927 

3181 
3386 
3579 
3766 
3945 

3201 
3404 
3598 
3784 
3962 

25 

26 
27 
28 
29 

3979 
4150 
4314 

4472 
4624 

3997 
4166 
4330 
4487 
4639 

4014 
4183 
4346 
4502 
4654 

4031 
4200 
4362 
4518 
4669 

4048 
4216 
4378 
4633 
4683 

4065 
4232 
4393 
4648 
4698 

4082 
4249 
4409 
4664 
4713 

4099 
4265 
4425 
4579 

4728 

4116 
4281 
4440 
4594 
4742 

4133 

4298 
4456 
4609 

4757 

30 

31 
32 
33 
34 

4771 
4914 
5051 
5185 
5315 

4786 
4928 
5065 
5198 
5328 

4800 
4942 
5079 
5211 
5340 

4814 
4955 
5092 
5224 
5353 

4829 
4969 
5105 
5237 
6366 

4843 
4983 
5119 
6250 
5378 

4867 
4997 
6132 
5263 
5391 

4871 
5011 
6145 
5276 
6403 

4886 
5024 
6159 
6289 
6416 

4900 
5038 
5172 
6302 

6428 

35 

36 
37 
38 
39 

5441 
5563 
5682 
5798 
5911 

5453 
5575 
5694 
5809 
5922 

5465 

5587 
5706 
5821 
5933 

5478 
5599 
5717 
5832 
5944 

6490 
6611 
6729 
6843 
6956 

5602 
5623 
5740 
6865 
5966 

6614 
5636 
6752 
5866 
6977 

5627 
5647 
5763 

6877 
6988 

6639 
5658 
5775 
5888 
5999 

6551 
6670 
6786 
6899 
6010 

40 

41 
42 
43 
44 

6021 
6128 
6232 
6335 
6435 

6031 
6138 
6243 
6345 
6444 

6042 
6149 
6253 
6355 
6454 

6053 
6160 
6263 
6365 
6464 

6064 
6170 
6274 
6375 
6474 

6076 
6180 
6284 
6386 
6484 

6086 
6191 
6294 
6396 
6493 

6096 
6201 
6304 
6405 
6603 

6107 
6212 
6314 
6415 
6513 

6117 
6222 
6326 
6425 
6522 

45 

46 
47 
48 
49 

6532 

6628 
6721 
6812 
6902 

6542 
6637 
6730 
6821 
6911 

6551 
6646 
6739 
6830 
6920 

6661 
6656 
6749 
6839 
6928 

6571 
6666 
6758 
6848 
6937 

6680 
6676 
6767 
7867 
6946 

6690 
6684 
6776 
6866 
6965 

6699 
6693 
6786 
6875 
6964 

6609 
6702 
6794 
6884 
6972 

6618 
6712 
6803 
6893 
6981 

50 
61 
52 
53 
54 

6990 
7076 
7160 
7243 
7324 

6998 
7084 
7168 
7251 
7332 

7007 
7093 
7177 
7259 
7340 

7016 
7101 
7186 
7267 
7348 

7024 
7110 
7193 
7276 
7356 

7033 
7118 
•7202 
7284 
7364 

7042 
7126 
7210 
7292 
7372 

7060 
7135 

7218 
7300 
7380 

7069 
7143 

7226 
7308 
7388 

7067 
7152 
7235 
7316 
7396 

LOGARITHMS 


453 


N   0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

55 

56 
57 
58 
59 

7404  7412 
7482  1  7490 
7559  1  7566 
7634  7642 
7709  7716 

7419 
7497 
7574 
7649 
7723 

7427 
7506 
7582 
7657 
7731 

7435 
7513 
7589 
7664 

7738 

7443 
7520 
7597 
7672 
7746 

7461 

7528 
7604 
7679 
7752 

7459 
7536 
7612 
7686 
7760 

7466 
7543 
7619 
7694 
7767 

7474 
7651 

7627 
7701 

7774 

60 
61 
62 
63 
64 

7782 
7853 
7924 
7993 
8062 

7789 
7860 
7931 
8000 
8069 

7796 
7868 
7938 
8007 
8075 

7803 
7875 
7945 
8014 
8082 

7810 
7882 
7952 
8021 
8089 

7818 
7889 
7959 

8028 
8096 

7826 
7896 
7966 
8036 
8102 

7832 
7903 
7973 
8041 
8109 

7839 
7910 
7980 
8048 
8116 

7846 
7917 
7987 
8065 
8122 

65 

66 
67 
68 
69 

8129 
8195 
8261 
8325 
8388 

8136 
8202 
8267 
8331 
8395 

8142 
8209 
8274 
8338 
8401 

8149 
8216 
8280 
8344 
8407 

8156 
8222 
8287 
8351 
8414 

8162 
8228 
8293 
8357 
8420 

8169 
8235 
8299 
8363 
8426 

8176 
8241 
8306 
8370 
8432 

8182 
8248 
8312 
8376 
8439 

8189 
8254 
8319 
8382 
8446 

70 

71 
72 
73 

74 

8451 
8513 
8573 
8633 
8692 

8457 
8519 
8579 
8639 
8698 

8463 
8526 
8585 
8645 
8704 

8470 
8631 
8691 
8661 
8710 

8476 
8637 
8597 
8667 
8716 

8482 
8543 
8603 
8663 
8722 

8488 
8549 
8609 
8669 

8727 

8494 
8556 
8615 
8675 
8733 

8500 
8561 
8621 
8681 
8739 

8506 
8567 
8627 
8686 
8746 

75 

76 
77 
78 
79 

8751 
8808 
8865 
8921 
8976 

8756 
8814 
8871 
8927 
8982 

8762 
8820 
8876 
8932 
8987 

8768 
8826 
8882 
8938 
8993 

8774 
8831 
8887 
8943 
8998 

8779 
8837 
8893 
8949 
9004 

8785 
8842 
8899 
8964 
9009 

8791 
8848 
8904 
8960 
9016 

8797 
8854 
8910 
8966 
9020 

8802 
8859 
8915 
8971 
9025 

80 
81 
82 
83 
84 

9031 
9085 
9138 
9191 
9243 

9036 
9090 
9143 
9196 
9248 

9042 
9096 
9149 
9201 
9253 

9047 
9101 
9164 
9206 
9268 

9063 
9106 
9159 
9212 
9263 

9058 
9112 
9165 
9217 
9269 

9063 
9117 
9170 
9222 
9274 

9069 
9122 
9176 
9227 
9279 

9074 
9128 
9180 
9232 
9284 

9079 
9133 
9186 
9238 
9289 

85 
86 
87 
88 
89 

9294 
9345 
9395 
9445 
9494 

9299 
9350 
9400 
9450 
9499 

9304 
9356 
9406 
9455 
9504 

9309 
9360 
9410 
9460 
9509 

9316 
9366 
9415 
94(55 
9513 

9320 
9370 
9420 

9469 
9518 

9326 
9376 
9426 
9474 
9623 

9330 
9380 
9430 
9479 

9528 

9336 
9385 
9436 
9484 
9533 

9340 
9390 
9440 
9489 
9638 

90 
91 
92 
93 
94 

9542 
9590 
9638 
9685 
9731 

9547 
9595 
9643 
9689 
9736 

9552 
9600 
9647 
9694 
9741 

9557 
9605 
9652 
9699 
9745 

9562 
9609 
9657 
9703 
9750 

9566 
9614 
9661 
9708 
9754 

9571 
9619 
9666 
9713 
9759 

9676 
9624 
9671 
9717 
9763 

9581 
9628 
9675 
9722 
9768 

9586 
9633 
9680 
9727 
9773 

95 

96 
97 
98 
99 

9777 
9823 
9868 
9912 
9956 

9782 
9827 
9872 
9917 
9961 

9786 
9832 
9877 
9921 
9965 

9791 
9836 
9881 
9926 
9969 

9795 
9841 
9886 
9930 
9974 

9800 
9845 
9890 
9934 
9978 

9805 
9850 
9894 
9939 
9983 

9809 
9864 
9899 
9943 

9987 

9814 
9859 
9903 
9948 
9991 

9818 
9863 
9908 
9962 
9996 

454  LOGARITHMS 

Pind  the  numbers  corresponding  to  the  following  logarithms: 

21.  1.3179.  26.  2.9900.           31.   1.5972.  36.  0.2468. 

22.  3.0146.  27.  0.1731.           32.    1.0011.  37.  0.1357. 

23.  0.7145.  28.  0.8974.           33.   2.7947.  38.  2.0246. 

24.  1.5983.  29.  0.9171.           34.    2.5432.  39.  1.1358. 

25.  2.0013.  30.  3.4015.           35..  0.5987.  40.  4.0478. 

202.  Products  and  powers  may  be  found  by  means  of  loga- 
rithms, as  shown  by  the  following  examples. 

Ex.  1.   Find  the  product  49  X  134  x  .071  x  349. 
Solution.     From  the  table, 

log  49  =  1.6902  or    49  =  10i-6902. 

log  134  =  2.1271  or  134  =  IO21271. 

log  .071  =  2.8513  or  .071  =  lO^-ssia. 

log  349  =  2.5428  or  349  =  102-5428. 

Since  powers  of  the  same  base  are  multiplied  by  adding  exponents, 

§  176,  we  have       49  x  134  x  .071  x  349  =  106-2i". 

Hence  log  (49  x  134  x  .071  x  349)  =  5.2114. 

The  number  corresponding  to  this  logarithm,  as  found  by  the 
method  used  in  Ex.  3  above,  is  162704.  By  actual  multiplication  the 
product  is  found  to  be  162698.914  or  162699  which  is  the  nearest 
approximation  without  decimals.  Hence  the  product  obtained  by 
means  of  logarithms  is  5  too  large.  This  is  an  error  of  about  s^^s  of 
the  actual  result  and  is  therefore  so  small  as  to  be  negligible. 

Ex.  2.   Find  (1.05)20. 

Solution.      Log  1.05  =  0.0212  or  100-0212  =  1.05. 
Hence  (1.05)2o  =  (100-0212)20  =  io(o-o2i2).ao  _  100.424^ 

or  log  (1.05)20  =  0.4240. 

Hence  (1.05)2o  is  the  number  corresponding  to  the  logarithm  0.4240. 


LOGARITHMS  455 

Since  logarithms  are  exponents  of  the  base  10,  it  follows  from 
the  laws  of  exponents  (see  §  198)  that 

(a)   The  logarithm  of  the  product  of  two  or  more  numbers  is 
the  sum  of  the  logarithms  of  the  numbers. 

(6)   TJie  logarithm  of  a  power  of  a  number  is  the  logarithm  of 
the  number  multiplied  by  the  index  of  the  power. 

That  is, 
/og  (a  '  b  '  c)  =  foga  +  /og  b  i-  log  c,  and  log  a"  =  n  log  a. 

EXERCISES 

By  means  of  the  logarithms  obtain  the  following  products 
and  powers : 

1.  243  X  76  X  .34.  7.  5.93  x  10.02.  13.  (49)«  x  19  x  21^. 

2.  823.68  X  370.  8.  4S6  x  3.45.  14.  .21084  x  (.53)2. 

3.  216.83  x  2.03.  9.  (.02)2  ^  (o.8).  15.  7.865  x  (.013)^. 

4.  572  X  (.71)2.  10    ((55)2  X  (^91)3^  ^Q  (g  75)3  y^  (j23y. 

5.  510  X  (9.1)^  11.  (84)2  ^  (75)3_  ^^  (1.46)2  x  (61.2)2. 

6.  43.71  x"  (21)2.  12.    (.960)2(49)2.  18.  (3.54)3  x  (29.3)2. 

19.    (4.132)2  x  (5.184)2.         20.   1946  x  398  x  .08. 

203.   Quotients  and  roots  may  be  found  by  means  of  logarithms, 
as  shown  by  the  following  examples. 

Ex.  1.   Divide  379  by  793. 

Solution.     From  the  table, 

log  379  =  2.5786  or  102-6786  =  379. 

log  793  =  2.8993  or  102-8998  =  793. 

Hence  by  the  law  of  exponents  for  division,  §  175, 

379  -^  793  =  102-6786-2.8998. 

Since  in  all  operations  with  logarithms  the  mantissa  is  positive,  write 
the  first  exponent  3.5786  —  1  and  then  subtract  2.8993. 

Hence  log  (379  -  793)  =.6793  -  1  =  1.6793. 

Hence  the  quotient  is  the  number  corresponding  to  this  logarithm. 


456  LOGARITHMS 


Ex.  2.   By  means  of  logarithms  approximate  V422  X  37^ 
By  the  methods  used  above  we  find 

log  (422  X  375)=  11.0874  or  W^-^^^*  =  42^  x  37s. 

11.0874 

Hence  v'422  x  37^  =(W^-os74^i=  10    ^     =103-6958. 


That  is,  log  V422  x  37«  =  3.6958. 

Hence  the   result   nought   is  the    number  corresponding  to  this 
logarithm. 

It  follows  from  the  laws  of  exponents  (see  §  198)  that 

(a)  TJie  logarithm  of  a  quotient  equals  the  logarithm  of  the 
dividend  minus  the  logarithm  of  the  divisor. 

(b)  TJie  logarithm  of  a  root  of  a  number  is  the  logarithm  of 
the  number  divided  by  the  index  of  the  root. 

That  is 

log ?  =  logfl  —  log  b  and  log  Va  =  -^^. 
D  n 

EXERCISES 

By  means  of   logarithms   approximate  the  following  quo- 
tients and  roots : 


1.  45.2-8.9.          4.  V196x256.  7.  -^15  x -v/67. 

2.  231.18 -.4.2.      5.   §?l>i^^.  8.  ^211x^34T. 

2v  .43  X  3.246 

3.  .04905 -^-.327.    e.   ■</69  - -^21.  ^'  (5184)U  (38124)i 


10.  (6.75)3 -(2.132)1  ^^     .3/l3^x.3Px4.31« 


11.-^/105-^^.  ^V71xV41xV5l 

12.  (91125)U(576)i  ^^  .    ,\      4^  x  .57«  x  42^~. 

13.  (3.040)3  -  (.0065)3.  *    A  ^^  ^  ■y/'M  x  V23 

14.  (29.3)UV(3:47)^.  ^^     /J^54>^j^282<J^\f 

15.  -v/39  X  -V^  x  </WJ.  '    W4'  X  -v^  X  (.003)  V* 


CHAPTER  XII 
PROGRESSIONS 

ARITHMETIC    PROGRESSIONS 

204.  An  arithmetic  progression  is  a  series  of  numbers,  such 
that  any  one  after  the  first  is  obtained  by  adding  a  fixed  num- 
ber to  the  preceding.  The  fixed  number  is  called  the  common 
difference. 

The  general  form  of  an  arithmetic  progression  is 

a,   a  +  d,   a-\-2d,   a-\-Zd,    •••, 

where  a  is  the  first  term  and  d  the  common  difference. 

E.g.  2,  5,  8, 11, 14,  •••  is  an  arithmetic  progression  in  which  2  is  the 
first  term  and  3  the  common  difference.  Written  in  the  general  form, 
it  would  be      o,  2  +  3,  2  +  2-3,  2  +  3-3,  2+4-3,   .... 

205.  If  there  are  n  terms  in  the  progression,  then  the  last 
term  is  a-\-(n  —  V)d.     Indicating  the  last  term  by  I,  we  have 

/=ff+(„-iy.  I 

An  arithmetic  progression  of  n  terms  would  then  be  written 
in  general  form,  thus, 

a,   a  +  d,   a-\-2d,     ..,   a+(/i-2)(/,  a+{n'-l)d. 

EXERCISES 

1.    Solve  I  for  each  letter  in  terms  of  all  the  others. 

In  each  of  the  following  find  the  value  of  the  letter  not  given,  and 
write  out  the  progression  in  each  case. 

cZ  =  2,  3.    I  d  =  5,  4.    J  71  =  15,         5.    I  n  =  31, 

n  =  l.  i?  =  43.  [z  =  15.  tz  =  91. 

457 


458  PROGRESSIONS 

ra  =  4,  (a  =  S,  fd  =  -5,  ra  =  ll, 

6.    |d  =  -3,      8.    |d  =  -5,    10.   |w  =  13,        12.    h  =  -39, 
[71  =  18.  [l  =  -S2.  [z  =  -63.  [cZ  =  -5. 

{a  =  —  5,  rc^  =  7,  fa  =  —  3,  ra  =  a;, 

d  =  4,  9.    Jn  =  8,         11.    \n  =  9,          13.    h  =  2/, 

n  =  7.  [1  =  24..  [l  =  ~27,  [n  =  z. 

206.  The  sum  of  an  arithmetic  progression  of  n  terms  may  be 
obtained  as  follows : 

Let  Sn  denote  the  sum  of  n  terms  of  the  progression.     Then, 

s«  =  a  +  [«  +  flf]  +  [a  +  2fi]+  ...  H-[a  +  (n-2)<Z]  +  [a  +  (n-l)c?].   (1) 

This  may  also  be  written,  reversing  the  order  of  the  terms,  thus, 

s,  =  [a+(n-  1)^]  +[a  +  (n-2)f/]+  ...  +  [a  +  2 J]  +  [a+  rf]  +  a.  (2) 

Adding  (1)  and  (2),  we  have 
2s„  =  [2a  +  (n  -  1)^]  +  [2  a  +  (n  -  2)rf  +  d} 

+  ...  +[2a+(n-2)cf +  <]  +  [2a+(n-l)rf]. 

The  expression  in   each  bracket  is   reducible  to  2  a  +  (w  —  l)rf, 
which  may  also  be  written  a  +[_a  +  (n  —  l)d']  =  a  +  ^,  by  §  205. 
Since  there  are  n  of  these  expressions,  each  a  +  /,  we  have 

2s„  =  n(a  +  I). 
Hence  s„  =  ^(a-\-/).  '  II 

This  formula  for  the  sum  of  n  terms  involves  a,  I,  and  w,  that 
is,  the  first  term,  the  last  term,  and  the  number  of  terms. 

207.  In  the  two  equations, 

l  =  a  +  {n  —  l)dy  '  I 

s  =  |(a  +  0,  II 

there  are  five  letters,  namely,  a,  d,  I,  n,  s.  If  any  three  of 
these  are  given,  the  equations  I  and  II  may  be  solved  simul- 
taneously to  find  the  other  two,  considered  as  the  unknowns. 


ARITHMETIC   PROGRESSIONS  459 

The  solution  of  problems  in  arithmetic  progression  by  means 
of  equations  I  and  II  is  illustrated  in  the  following  examples : 

Ex.1.   Given  71  =  11,   Z  =  23,   s=143.     Find  a  and  d 

Substituting  the  given  values  iu  I  and  II, 

23  =  a+(ll-l)c?.  (1) 

143  =  V(«  +  23).  (2) 

From  (2),  a  =  3,  which  in  (1)  gives  d  =  2. 

Ex.  2.   Given  d^A,   n=5,   s  =  75.     Eind  a  and  I 
From  I  and  II,  /  =  a  +  (5  -  1)4,  (1) 

75=|(a  +  0-  (2) 

Solving  (1)  and  (2)  simultaneously,  we  have  a  =  7,  Z  =  23. 

Ex.  3.   Given  c?  =  4,   I  =  35,   8  =  161.     Find  a  and  n. 
From  I  and  II,  35  =  a  +  (n  -  1)4,  (1) 

161=|(a  +  35).  (2) 

From  (1)  a  =  39  -  4  n, 

which  in  (2)  gives  161  =  ^(74  -  4  n)  =  37  n  -  2  n*. 

Hence  *  n  =  -2;^,  or  7.  • 

Since  an  arithmetic  progression  must  have  an  integral  number  of 
terms,  only  the  second  value  is  applicable  to  this  problem. 

Ex.  4.   Given  d  =  2,   Z  =  11,   s  =  35.     Find  a  and  n. 
Substituting  in  I  and  II,  and  solving  for  a  and  n,  we  have 
a  =  3,   n  =  5,   and  a  =  —  1,   n  =  7. 

Hence  there  are  two  progressions, 

-  1,  1,  3,  5,  7,  9,  11, 
and  3,  5,  7,  9,  11, 

each  of  which  satis^es  tlie  given  conditions. 


460  PKOGRESSIONS 

EXERCISES 

In  each  of  the  following  obtain  the  values  of  the  two  letters 
not  given. 

If  fractional  or  negative  values  of  n  are  obtained,  such  a  result  in- 
dicates that  the  problem  is  impossible.  This  is  also  the  case  if  an 
imaginary  value  is  obtaiped  for  any  letter.  In  each  exercise  interpret 
all  the  values  found. 

rs  =  96,  rs  =  88,  \d=-l, 

1.  h  =  19,       4.    |z=-7,      7.    ]ri=41,         10. 

rs=34,  rn=18,  U  =  30,  rs=7, 

2.  \l  =  U,       5.      a=4,  8.      s=162,  11.    \d=l^, 
[d=S.                 [1  =  13.               [n=9.  [1=7. 

\ 

3.  \l  =  27,       6.    ]a=7,  9.    ]n=10,  12.    \d=3, 
U  =  187.            is  =  14.               is=120.  [1  =  4.. 

In  each  of  the  following  call  the  two  letters  specified  the  unknowns 
and  solve  for  their  values  in  terms  of  the  remaining  three  letters 
supposed  to  be  known. 

13.  a,  d.  15.    a,  n.      ,    17.    d,  I.      '   19.   d,  s.  21.    Z,  s. 

14.  a,  I.  16.    a,  s.  18.    d,  n.        20.    I,  n.  22.    n,  s. 

208.  Arithmetic  means.  The  terms  between  the  Brst  and  the 
last  of  an  arithmetic  progression  are  called  arithmetic  means. 

Thus,  in  2,  5,  8,  11,  14,  17,  the  four  arithmetic  means  between  2 
and  17  are  5,  8,  11,  14. 

If  the  first  and  the  last  terms  and  the  number  of  arith- 
metic means  between  them  are  given,  then  these  means  can  be 
found. 

For  we  have  given  a,  I,  and  n.  Hence  d  can  be  found  and  the 
whole  series  constructed. 


ARITHMETIC   PROGRESSIONS  461 

Example.     Insert  7  arithmetic  means  between  3  and  19. 

In  this  progression  a  =  3,  I  =  19,  and  n  =  9. 
Hence  from  l  =  a  +  (n  —  l)d  we  find  d  =  2  and  the  required  means 
are  5,  7,  9,  11, 13,  15,  17. 

209.  The  case  of  one  arithmetic  mean  is  important.  Let  A 
be  the  arithmetic  mean  between  a  and  I.  Since  a,  Ay  I  are  in 
arithmetic  progression,  we  have  A  =  a-\-d,  and  l=:A-{-d. 
Hence  A-l=^a-A 

or  >»=^'-  in 

EXERCISES  AND  PROBLEMS 

1.  Insert  5  arithmetic  means  between  5  and  —  7.  . 

2.  Insert  3  arithmetic  means  between  —  2  and  12. 

3.  Insert  8  arithmetic  means  between  —  3  and  —  5. 

4.  Insert  5  arithmetic  means  betw^een  —  11  and  40. 

5.  Insert  15  arithmetic  means  between  1  and  2. 

6.  Insert  9  arithmetic  means  between  2|  and  —  1^. 

7.  Find  the  arithmetic  mean  between  3  and  17. 

8.  Find  the  arithmetic  mean  between  —  4  and  16. 

9.  Find   the  tenth   and   eighteenth   terms    of  the   series 
4,  7,  10,  .... 

10.  Find  the  fifteenth  and  twentieth  terms  of  the  series 
-8,-4,0,.... 

11.  The  fifth  term  of  an  arithmetic  progression  is  13  and 
the  thirtieth  term  is  49.     Find  the  common  difference. 

12.  Find  the  sum  of  all  the  integers  from  1  to  100. 

13.  Find  the  sum  of  all  the  odd  integers  between  0  and  200. 

14.  Find  the  sum  of  all  integers  divisible  by  6  between  1 
and  500. 

15.  Show  that  1  +  3  4-  5  +  •••  -{-n  =  k^  where  k  is  the  number 
of  terms. 


•162  PROGRESSIONS 

16.  In  a  potato  race  40  potatoes  are  placed  in  a  straight  line 
one  yard  apart,  the  first  potato  being  two  yards  from  the 
basket.  How  far  must  a  contestant  travel  in  bringing  them 
to  the  basket  one  at  a  time  ? 

17.  There  are  three  numbers  in  arithmetic  progression  whose 
sum  is  15.  The  product  of  the  first  and  last  is  3^  times  the 
second.     Find  the  numbers. 

18.  There  are  four  numbers  in  arithmetic  progression  whose 
sum  is  20  and  the  sum  of  whose  squares  is  120.  Find  the 
numbers. 

19.  If  a  body  falls  from  rest  16.08  feet  the  first  second, 
48.24  feet  the  second  second,  80.40  the  third,  etc.,  how  far  will 
it  fall  in  10  seconds  ?  15  seconds  ?  t  seconds  ? 

20.  According  to  the  law  indicated  in  problem  19  in  how 
many  seconds  will  a  body  fall  1000  feet  ?  s  feet  ? 

If  a  body  is  thrown  downward  with  a  velocity  of  v^  feet  per  second, 
then  the  distance,  s,  it  will  fall  in  t  seconds  is  VqI  feet  plus  the  distance 
it  would  fall  if  starting  from  rest. 

That  is,  5  =  1^0^  +  ^9*"^^  where  g  =  32.16. 

21.  In  what  time  will  a  body  fall  1000  feet  if  thrown  down- 
ward with  a  velocity  of  20  feet  per  second  ? 

22.  With  what  velocity  must  a  body  be  thrown  downward 
in  order  that  it  shall  fall  360  feet  in  3  seconds  ? 

23.  A  stone  is  dropped  into  a  well,  and  the  sound   of  its 
striking  the  bottom  is  heard  in  3  seconds.     How  deep  is  the   * 
well  if  sound  travels  1080  feet  per  second  ? 

A  body  thrown  upward  with  a  certain  velocity  will  rise  as  far  as  it 
would  have  to  fall  to  acquire  this  velocity.  The  velocity  (neglecting 
the  resistance  of  the  atmosphere)  of  a  body  starting  from  rest  is  gi 
where  g  =  32.16  and  t  is  the  number  of  seconds. 

24.  A  rifle  bullet  is  shot  directly  upward  with  a  velocity  of 
2000  feet  per  second.  How  high  will  it  rise,  and  how  long  be- 
fore it  will  reach  the  ground  ? 


GEOMETRIC   PROGRESSIONS  463 

25.  From  a  balloon  5800  feet  above  the  earth,  a  body  is 
thrown  downward  with  a  velocity  of  40  feet  per  second.  In 
how  many  seconds  will  it  reach  the  ground  ? 

26.  If  in  Problem  25  the  body  is  thrown  upward  at  the  rate 
of  40  feet  per  second,  how  long  before  it  will  reach  the  ground  ? 

GEOMETRIC   PROGRESSIONS 

210.  A  geometric  progression  is  a  series  of  numbers  in  which 
any  term  after  the  first  is  obtained  by  multiplying  the  pre- 
ceding term  by  a  fixed  number,  called  the  common  ratio. 

The  general  form  of  a  geometric  progression  is 

a,   an,   ar,   ar^,  •••,  ar""\ 

in  which  a  is  the  first  term,  r  the  constant  multiplier,  or  com- 
mon ratio,  and  n  the  number  of  terms. 

E.g.  3,  6,  12,  24,  48,  is  a  geometric  progression  in  which  3  is  the 
first  term,  2  is  the  common  ratio,  and  5  is  the  number  of  terms. 

Written  in  the  general  form  it  would  be  3,  3  •  2,  3  .  2^,  3  •  2^,  3  •  2*. 

211.  If  I  is  the  last  or  nth  term  of  the  series,  then 

/  =  ar"-\  I 

If  any  three  of  the  four  letters  in  I  are  given,  the  remaining 
one  may  be  found  by  solving  this  equation. 

EXERCISES 

In  each  of  the  following  find  the  value  of  the  letter  not  given : 

rz=32, 

1.  {r  =  3,         4.    \r  =  -2,       7.    ]r  =  f,  10.       r=-2, 

[n  =  6. 

2.  ^r  =  2,         5.    U  =  -2,       8.    lr  =  h  11.       r=-f, 

U  =  7. 

fa  =  3,. 

3.  \r=:-3,     6.    \r  =  2,  9.    ^r  =  -f,      12.    |r  =  2, 

U-1536. 


464  PROGRESSIONS 

212.  The  sum  of  n  terms  of  a  geometric  expression  may  be 
found  as  follows : 

If  Sn  denotes  the  sum  of  n  terms,  then 

s„  =  a+ ar  +  ar2+ ... +  ar"-2+ar"-^  (1) 

Multiplying  both  members  of  (1)  by  r,  we  have 
« 

rsn  =  ar  +  ar'^  +  ar^  +  ...  +  ar^-'^  +  ar".  (2) 

Subtracting  (1)  from  (2),  and  canceling  terras,  we  have 

rSn  -  Sn  =  ar"^  —  a.  (3) 

Solving  (3)  for  s„  we  have 

an"  —  a      a(r"  —  1)  tt 

r  —  1  r~l 

This  formula  for  the  sum  of  n  terms  of  a  geometric  series 
involves  only  a,  r,  and  n. 

Since  ar"  =  r  •  ar"~^  —  r  -I,  s"  may  also  be  written  : 

r—1       1—r 

This  formula  involves  only  r,  I,  and  a. 

213.  Prom  equations  I  and  II  or  I  and  III  any  two  of  the 

numbers  a,  I,  r,  s,  and  n  can  be  found  when  the  other  three  are 
given,  as  in  the  following  examples. 

Ex.  1.   Given  n  =  T,  r  =  2,  s  =  381.    Find  a  and  I 

From  I  and  III,  l  =  a-2^=64:a,  (1 ) 

381  =  ^Isz±  =  21 -a.  (2) 

2  —  1 

Substituting  Z  =  61  a  in  (2),  we  obtain  a  =  3,  and  I  =  192. 

Ex.  2.    Given  a  =  -  3,  Z  =  -  243,  s  =  — 183.    Find  r  and  n. 
From  I  and  III,  -  243  =  (-  3)r«-i,  (1) 

_183  =  Il243il±i.  (2) 

r  —  1 


GEOMETRIC   PROGRESSIONS 


465 


From  (2)  r  =  -  3. 

From  (1)  81  =  (-3)"-!. 

Since  (—  3)*  =  81,  we  have  n  —  1  =  4  or  n  : 


(3) 
(4) 


EXERCISES 

1.  Solve  II  for  a  in  terms  of  the  remaining  letters. 

2.  Solve  III  for  each  letter  in  terms  of  the  remaining  letters. 
In  each  of  the  following  find  the  terms  represented  by  the 

interrogation  points. 


3. 


fa  =  l, 

s  =  635, 

s  =  13, 

r^=-i^ 

>  =  3, 

n  =  5, 

4.    \ 

5. 

6.    i 

71  =  5, 

.s  =  ? 

.a  =  ? 

a  =  ? 

['•=1- 

fa  =  l, 

>  =  i 

\r-=h 

[«=f. 

«=l-i 

?i  =  5, 

n  =  S, 

n=r, 

^=-H, 

8. 

^  =  1296, 

9. 

s  =  1050f, 

10. 

'=». 

r  =  ? 

a=? 

l=? 
a  =  ? 

r  =  ? 

,s  =  ? 

214.  Geometric  means.  The  terms  between  the  first  and  the 
last  of  a  geometric  progression  are  called  geometric  means. 

Thus  in  3,  6,  12,  24,  48,  three  geometric  means  between  3  and  48 
are  6,  12  and  24. 

If  the  first  term,  tho  last  term,  and  the  number  of  geometric 
means  are  given,  the  ratio  may  be  found  from  I,  and  then  the 
means  may  be  inserted. 

Example.    Insert  4  geometric  means  between  2  and  64. 
We  have  given  a  ^  2,  Z  =  64,  n  =  4  +  2  =  6,  to  find  r. 
From  I,  64  =  2  •  r6  -1  or  r^  =  32  and  r  =  2. 

Hence,  the  series  is  2, 4,  8,  16,  32,  64. 

215.  The  case  of  one  geometric  mean  is  important.     If  G  is 

C       h 
the  geometric  mean  between*  a  and  h,  we  have  —  =  — • 

a      G 


Hence, 


^  =  V^. 


466  PROGRESSIONS 

216.  Problem.  In  attempting  to  reduce  f  to  a  decimal,  we 
find  by  division  .666  •••,  the  dots  indicating  that  the  process 
goes  on  indefinitely. 

Conversely,  we  see  that  .666  -"  =j\-\-  ^^  +  ^-^^  +  •  •  •,  that 
is,  a  geometric  progression  in  which  a  =  -j^,  r  =  ^^,  and  n  is 
not  fixed  but  goes  on  increasing  indefinitely. 

At  n  grows  large,  I  grows  small,  and  by  taking  n  sufficiently  large, 
I  can  be  made  as  small  as  we  please.  Hence  formula  III,  §  212,  is  to 
be  interpreted  in  this  case  as  follows : 

_a~rl_10      10_6-/ 

10 
in  which  I  grows  small  indefinitely  as  n  increases  indefinitely,  so  that 
by  taking  n  large  enough  s„  can  be  made  to  differ  as  little  as  we  please 
from  1=^  =  ^  =  2. 
9         9      3 
In  this  case  we  say  s„  approaches  J  as  a  limit  as  n  increases 
indefinitely. 

Observe  that  this  interpretation  can  apply  only  when  the 
constant  multiplier  r  is  a  proper  fraction. 

EXERCISES  AND  PROBLEMS 

1.  Insert  5  geometric  means  between  2  and  128. 

2.  Insert  7  geometric  means  between  1  and  2^6' 

3.  Find  the  geometric  mean  between  8  and  18. 

4.  Find  the  geometric  mean  between  -^  and  ^. 

6.  Find  the  fraction  which  is  the  limit  of  .333  •••. 

6.  Find  the  fraction  which  is  the  limit  of  .1666  ••*. 

7.  Find  the  fraction  which  is  the  limit  of  .08333  •••. 

8.  Find  the  13th  term  of  -J/,  4,  -3  •••. 

9.  Find  the  sum  of  15  terms  of  the  series  —  243,  81,  —  27  •••. 
10.  Find  the  limit  of  the  sum   ■|H-|  +  i  +  i+  •••,  as  the 

number  of  terms  increases  indefinitely. 


GEOMETRIC   PROGRESSIONS  467 


Given 

Find 

Given 

Find 

11.   a,  r,  n 

l,s 

15.   a,  n,  I 

s,r 

12.    a,  ?•,  s 

I 

16.    r,  71,  I 

Sf  a 

13.   r,  n,  s 

I,  a 

17.    r,  I,  s 

a 

14.    a,  r,  I 

s 

18.   a,  I,  s 

r 

19.  The  product  of  three  terms  of  a  geometric  progression 
is  1000.     Find  the  second  term. 

20.  Four  numbers  are  in  geometric  progression.  The  sum 
of  the  second  and  third  is  18,  and  the  sum  of  the  first  and 
fourth  is  27.     Find  the  numbers. 

21.  Find  an  arithmetic  progression  whose  first  term  is  1 
and  whose  first,  second,  fifth,  and  fourteenth  terms  are  in  geo- 
metric progression. 

22.  Three  numbers  whose  sum  is  27  are  in  arithmetic  pro- 
gression.    If  1  is  added  to  the  first,  3  to  the  second,  and  11  to  . 
the  third  the  sums  will  be  in  geometric  progression.     Find 
the  numbers. 

23.  To  find  the  compound  interest  when  the  principal,  the 
rate  of  interest,  and  the  time  are  given. 

Solution.  Let  p  equal  the  number  of  dollars  invested,  r  the  rate  of 
per  cent  of  interest,  t  the  number  of  years,  and  a  the  amount  at  the 
end  of  t  years. 

Then      a  =  p(l  +  r)  at  the  end  of  one  year. 

a=p(l  +  r)(l  +  r)  =  p(l  +  ry  at  the  end  of  two  years. 

and  a  =  jt)  (1  +  r)'  at  the  end  of  t  years. 

That  is,  the  amount  for  t  years  is  the  last  term  of  a  geometric  pro- 
gression in  which  p  is  the  first  term,  1  +  r  is  the  ratio,  and  «  +  1  is  the 
number  of  terms. 

24.  Show  how  to  modify  the  solution  given  under  problem 
23  when  the  interest  is  compounded  semiannually ;  quarterly. 

25.  Solve  the  equation  a  =p  (1  +  rj  for  p  and  for  r. 


468  PROGRESSIONS 

26.  Solve  a  =p (1  +  r)*  for  t. 

Solution,     log  a  —  logjo(l  +  r)'  =  logjt?  +  log  (1  +  r)* 

=  log  »  +  Hog  (1  +  r).    (See  §  202.)     Hence  t  =  l2gj«L=lM£. 
^  log  (1  +  r) 

27.  At  what  rate  of  interest  compounded  annually  will 
$1200  amount  to  $1800  in  12  years? 

28.  At  what  rate  of  interest  compounded  semiannually  will 
a  sum  double  itself  in  20  years  ?  in  15  years  ?  in  10  years  ? 

29.  In  what  time  will  $8000  amount  to  $13,500,  the  rate 
of  interest  being  3^  %  compounded  annually  ? 

30.  In  what  time  will  a  sum  double  itself  at  3  %,  4  %,  5  %, 
compounded  semiannually  ? 

The  present  value  of  a  debt  due  at  some  future  time  is  a  sum  such 
that,  if  invested  at  compound  interest,  the  amount  at  the  end  of  the 
time  will  equal  the  debt. 

31.  What  is  the  present  value  of   $2500  due  in  4  years, 
*  money  being  worth  3|^  %  interest  compounded  semiannuallj^  ? 

32.  A  man  bequeathed  $  50,000  to  his  daughter,  payable  on 
her  twenty-fifth  birthday,  with  the  provision  that  the  present 
worth  of  the  bequest  should  be  paid  in  case  she  married  before 
that  time.  If  she  married  at  21,  how  much  would  she  receive, 
interest  being  4  %  per  annum  and  compounded  quarterly  ? 

33.  What  is  the  rate  of  interest  if  the  present  worth  of 
$24,000  due  in  7  years  is  $19,500  ? 

34.  In  how  many  years  is  $  5000  due  if  its  present  worth 
is  $3500,  the  rate  of  interest  being  3J%  compounded  annually? 

HARMONIC   PROGRESSIONS 
217.   A  harmonic  progression   is   a   series   whose  terms   are 
the  reciprocals  of  the  corresponding  terms  of  an  arithmetic 
progression. 

E.g.  1,  -  -,  -,  -  •••  is  a  harmonic  progression  whose  terms  are  the 
reciprocals  of  the  terms  of  the  arithmetic  progression  1,  3,  5,  7,  9  •••. 


HARMONIC   PROGRESSIONS  469 

The  name  harmonic  is  given  to  such  a  series  because  musical  strings 
of  uniform  size  and  tension,  whose  lengths  are  the  reciprocals  of  the 

positive  integers,  i.e.  1,  -,  -  -  •••,  vibrate  in  harmony. 

The  general  form  of  the  harmonic  progression  is 

1   _1 1_  1  J 

a' a  +  d'a-\-2d'  '"'    a  +  {n-l)d 

It  follows  that  if  a,  h,  c,  d,  e,  ...  are  in  harmonic  progression, 

then  -,  -,  -,-,-,  •  •  •  are  in  arithmetic  progression.     Hence,  all 
a  h  c  d  e 

questions  pertaining  to  a  harmonic  progression  are  best   an- 
swered by  first  converting  it  into  an  arithmetic  progression. 

218.  Harmonic  means.  The  terms  between  the  first  and  the 
last  of  a  harmonic  progression  are  called  harmonic  means  be- 
tween them. 

Example.     Insert  five  harmonic  means  between  30  and  3. 

This  is  done  by  inserting  five  arithmetic  means  between  -^  and  \. 
By  the  method  of  §  208  the  arithmetic  series  is  found  to  be  ^,  ^,  j^, 
kh  TTo.  6^.  i     Hence,  the  harmonic  series  is  30,  12,  Y,  f £,  ¥»  ^h  ^^ 

219.  The   case   of  a  single   harmonic   mean  is  important. 

Let  a,  H,  I  be  in  harmonic  progression.     Then  \  — ,  -  are  in 
arithmetic  progression.  ^ 

Hence,  by  §  209,  i  =  i^  or  «  =  ||^. 

220.  The  arithmetic,  geometric,  and  harmonic  means  be- 
tween a  and  I  are  related  as  follows : 

"We  have  seen 
Hence, 


A  = 

a  +  l 
2    ' 

G  = 

--  Val,  H  = 

2al 

a  +  l 

A 

2 

-^al 

_a  +  l 
2al 

■ 

A 

G'' 

■h- 

A  _ 
G 

G 
H 

Therefore, 

That  i^,  (r  is  a  mean  proportional  between  A  and  //.     See  §  172. 


470  PROGRESSIONS 

EXERCISES  AND  PROBLEMS 

1.  Insert  three  harmonic  means  between  22  and  11. 

2.  Insert  six  harmonic  means  between  i  and  ^^-. 

3.  The  first  term  of  a  harmonic  progression  is  ^  and  the 
tenth  term  is  2^-     Find  the  intervening  terms. 

4.  Two  consecutive  terms  of  a  harmonic  progression  are  5 
and  6.  Eind  the  next  two  terms  and  also  the  two  preceding 
terms. 

5.  If  a,  b,  c  are  in  harmonic  progression,  show  that 
a-T-c  =  (a  —  b)-i-(b  —  c). 

6.  rind  the  arithmetic,  geometric,  and  harmonic  means 
between : 

(a)  16  and  36  :  (b)  m  -\-  n  and  m  —  n :  (c) •  and . 

7n-\-n          m  —  n 

7.  The  harmonic  mean  between  two  numbers  exceeds  their 
arithmetic  mean  by  7,  and  one  number  is  three  times  the  other. 
Find  the  numbers. 

8.  If  X,  y,  and  z  are  in  arithmetic  progression,  show  that 
mx,  my,  and  mz  are  also  in  arithmetic  progression. 

9.  X,  y,  and  z  being  in  harmonic   progression,  show  that 
^ ,  and are  in  harmonic  progression, 


x-{-y-^z  x-\-y-\-z  x-\-y-\-z 

and  also  that     ^    ,  -^— ,  and  — —  are  in  harmonic  progression. 

y  +  z  x  +  z  x+y 

10.  The  sum  of  three  numbers  in  harmonic  progression  is  3, 
and  the  first  is  double  the  third.     Find  the  numbers. 

11.  The  geometric  mean  between  two  numbers  is  ^  and  the 
harmonic  mean  is  \.     Find  the  numbers. 

12.  Insert  n  harmonic  means  between  the  numbers  a  and  b. 


CHAPTER   XIII 

THE   BINOMIAL  FORMULA 

221.   In  Chapter  II  the  following  products  were  obtained : 
(a  +  &)2  =  a2  +  2a&  +  &2. 
(a  +  &)8  ;=  a^  +  3  a%  +  3  aZ»2  +  js. 
(a  +  by  =  a*  +  4  aSfe  +  6  a^b'^  +  4  ai^  -f  &4. 
(a  +  ?>)«  =  a^  -\-5a^b  +  10  a^i^  +  lo  a%^  +oab*-\-  h\ 

By  a  study  of  these  the  following  facts  may  be  observed : 

1.  Each  product  has  one  term  more  than  the  number  of  units  in 
the  exponent  of  the  binomial. 

2.  The  exponent  of  a  in  the^r*^  term  is  the  same  as  the  exponent 
of  the  binomial,  and  diminishes  by  unity  in  each  succeeding  term. 

The  exponent  of  h  in  the  last  term  is  the  same  as  the  exponent  of 
the  binomial,  and  diminishes  by  unity  in  e'dch  preceding  term. 

3.  The  sum  of  the  exponents  in  each  term  is  equal  to  the  expo- 
nent of  the  binomial. 

4.  The  coefficient  of  the  first  term  is  unity ;  of  the  second  term, 
the  same  as  the  exponent  of  the  binomial ;  and  the  coefficient  of 
any  other  term  may  be  found  by  nmltiplying  the  coefficient  of  the 
next  preceding  term  by  the  exponent  of  a  in  that  term  and  dividing 
this  product  by  a  number  one  greater  than  the  exponent  of  b  in 
that  term. 

5.  The  coefficients  of  any  pair  of  terms  equally  distant  from  the 
ends  are  equal. 

Statements  2  and  4  form  a  rule  for  writing  out  any  power 
of  a  binomial  np  to  the  fifth.     Let  us  find  (a  +  by. 

471 


472  THE   BINOMIAL  FORMULA 

Multiplying  (a  +  6)  ^  by  a  +  &,  we  have 

(a  +  hy(a  +  b)  =  a^  +  oa%  +  10  a%^  +  10  a%^  +    5  a%*  +     ab^ 
a^b  +    5  g'^^^  +  10  a^/^a  +  10  a^h^  +  pgh^  ^  b^ 

Hence  (a  +  i)^  =  a^  +  6  a^Z^  +  15  a^b^  +  20  a^^^  +  15  a%^  ^  Q  ab^  +  b^ 
From  this  it  is  seen  that  the  rule  holds  also  for  (a  +  by. 

PROOF  BY  MATHEMATICAL   INDUCTION 
222.   A   proof   that   the   above   rule   holds   for   all  positive 

integral  2)oivers  of  a  binomial  may  be  made  as  follows : 

First  step.     Write  out  the  product  as  it  would  be  for  the  nth 

power  on  the  supposition  that  the  rule  holds. 

Then  the  first  term  would  be  a"  and  the  last  term  5".     The  second 

terms  from  the  ends  would  be  na'^~^b  and  nab^-\     The  third  terms 

from  the  ends  would  be  ^(^~^)a»-262  and  ^^'^  ~  ^)  a^ft^-^.     The 

fourth  terms  from  the  ends  would  be 

n{n-l){n-2)        ,,    ^^  ^(n  -  l)(n  -  2)    3j,._3 
1-2-3  1.2.3  ' 

and  so  on,  giving  by  the  hypothesis,  ^ 

Second  step.  Multiply  this  expression  by  a  +  &  and  see  if 
the  result  can  be  so  arranged  as  to  conform  to  the  same  rule. 
Then,  (a  +  by  {a  +  b) 

=  ««+!  +  na«&  +  "^^~^^a"-^Z>2  +  •  • .  +  naV-i  +  «&« 

a"6  +  na"-i&2  +  . . .  +  ^p^  a%^-^  +  na&«  +  &«+i. 

Hence  adding, 
(a  +  by+^  =  a"+i  +  (n  +  !)«"&  +  f^^!""  ^^  +  nl  a^'W  +  •  •  • 

+  Tn  +  ^('|~^^"|  a%^-^  +  (n  +  l)aJ'»  +  &«+i. 

Combining  the  terms  in  brackets,  we  have, 

(a  +  &)«+!  =  an+i  +  (n  +  l)a"Z>  +  I^^^til^an-iftS  _^  . . . 


PROOF   BY   INDUCTION  478 

The  last  result  shows  that  the  rule  holds  for  (a-f  ft)**"^^  if  it 
holds  for  (a  -\-  by.  That  is,  if  the  rule  holds  for  any  positive  inte- 
gral exponent,  it  holds  for  the  next  higher  integer. 

Third  step.  It  was  found  above  by  actual  multiplication  that 
the  rule  does  hold  for  (a  +  ^Y-  Hence  by  the  above  argument 
we  know  that  the  rule  holds  for  (a  +  by. 

Moreover,  since  we  now  know  that  the  rule  holds  for  (a  +  hy,  we 
conclude  by  the  same  argument  that  it  holds  for  (a  +  ly,  and  if  for 
(a  +  hy,  then  for  (a  +  &)*,  and  so  on. 

Since  this  process  of  extending  to  higher  powers  can  be  car- 
ried on  indefinitely,  we  conclude  that  the  five  statements  in 
§  221  hold  for  all  positive  integral  powers  of  a  binomial. 

The  essence  of  this  proof  by  mathematical  induction  consists 
in  applying  the  supposed  rule  to  the  ni\\  power  and  finding 
that  the  rule  does  hold  for  the  (n  +l)th  power  if  it  holds  for 
the  nth  poiver. 

223.  The  general  term.  According  to  the  rule  now  known  to 
hold  for  any  positive  integral  exponent,  we  may  write  as  many 
terms  of  the  expansion  of  (a  +  by  as  may  be  desired,  thus : 

1  • « 

/?(/>-!)(/> -2)  ,.3  ,  n(n-l)in-2)(n-3)  ^_^.,  _  ^  j 
^        1.23  ^  12. 3. 4  ^       * 

.    From  this  result,  called  the  binomial  formula,  we  see : 

(1)  The  exponent  of  h  in  any  term  is  one  less  than  the  number  of 
that  term,  and  the  exponent  of  a  is  n  minus  the  exponent  of  b.  Hence 
the  exponent  of  b  in  the  (k  +  l)st  term  is  k,  and  that  of  a  is  n  —  A:. 

(2)  In  the  coefficient  of  any  term  the  last  factor  in  the  denominator 
is  the  same  as  the  exponent  of  b  in  that  term,  and  the  last  factor  in 
the  numerator  is  one  more  than  the  exponent  of  a. 

Hence  the  (k  -f  l)st  term,  which  is  called  the  general  term  is 

;,(;,  _  !)(;,  _  2)(„  -  3)  .  ^  >  (p  -  k  +  1)^.,,^,  IJ 

1 . 2  .  3  . 4  .  5  . . .  )t 


474  THE   BINOMIAL  FORMULA 

224.  The  process  of  writing  out  the  power  of  a  binomial  is 
called  expanding  the  binomial,  and  the  result  is  called  the  ex- 
pansion of  the  binomiaL 

Ex.  1.   Expand  {x  —  yy. 
In  this  case  a  =  x,  h  =  —  y,  n  =  4:. 
Hence  substituting  in  formula  I, 

(x-yy  =  x^  +  ^x%-y)  +  ^('^-^yx%-yy+^(^-^](^-^)x(-y)^ 

4C4-l)(4-2)(4-3) 

2.3.4  ^^^  ^  ^ 

=  :,4  _  4  ^8^  +  i^^2^2  _  ^^xy^  +  ^^;^y''  (2) 

Hence   {x  -  yy  =  x^- 4  x^y  +  Q  x^y^ -4:  xy^  +  y\  (3) 

Notice  that  this  is  precisely  the  same  as  the  expansion  of  (x  +  y)* 
except  that  every  other  term  beginning  with  the  second  is  negative. 

Ex.  2.   Expand  (1-2  y)\ 
Here  a  ==  1,  6  =  —  2  y,  n  =  5. 

Since  the  coefficients  in  the  expansion  of  (a  +  h)^  are  1, 5, 10, 10,  5, 1, 
we  write  at  once, 

(1  -  2  ?/)5  =  P  +  5  .  14 .  (-  2  2/)  +  10  .  18 .  (-  2  ?/)2 

+  10.12.(-23/)3  +  5.1.(-2  3/)'*+  (-2^)5 
=  1  -  10  ?/  +  40  2/2  -  80  2/3  ^  80  y^  -  32  y^ 

Ex.3.   Expand  (-  +  !)'. 

Remembering  the  coefficients  just  given,  we  write  at  once, 

x^     'dx*      9  a;3     27a:2      81  x      243* 


EXPANSION  OF  BINOMIALS  475 

In    a    similar    manner  any  positive   integral   power   of   a 
binomial  may  be  written. 

Ex.  4.   Write  the  sixth  term  in  the  expansion  of  {x—2  yf^ 
without  computing  any  other  term. 

From  II,  §  223,  we  know  the  (k  +  l)st  term  for  the  nth  power  of 

<'  +  *'--«'y'     n(n-l)(n-2)-(,.-t+ !)„._.,. 
2-3-4-i 

In  this   case  a  =  x,   h  =  —  2  y,   n  =  10,   k  +  1  =  Q.     Hence  ^  =  5. 
Substituting  these  particular  values,  we  have 

ioao-i)ao-2)...ao-5+i)^,,.,.  ^  ., 
2.3.4.5  .^    ^^ 

^10.9.8.7.6^,       ,^ 
2.3.4.5     ^     ^^ 

=  -  32  .  252  x^y^  =  -  8064  xY- 


EXERCISES 

1.   Make  a  list  of  the  coefficients  for  each  power  of  a  binomial 
from  the  2d  to  the  10th. 


Expand  the  following : 
2.    (x  —  yy.  9.    (x^  —  y^y. 


17. 


^_yy 
y    ^J 


3.  (2aj  +  3)«.  10.   {x-^^y-y, 

4.  (^x  +  2yy.        11-    (— ^)^-  18.    r^-2/V^Y. 

12.  (x-^v^\  V2/"         y 


5.    (3  +  2/)^ 


12.    {x  +  yy. 


13.  (m-7iy«.  fy^_,^ly\ 

6.    (3^  +  2/)'.  14.    (r^  +  sO^  '    \Vn'     ^rtj  ' 

7.  (a.-2/«)«.         15.  (c-^-d-^y,      ^^  fcJ/^_<lA\ 

8.    (a^-2/y.  16.    (Va-V5)^  '    Vvd^        ^7 

21.    (2a^x-^-3by-y.  22.    (3  xy-^  -  x-^yy. 


476  THE  BINOMIAL  FORMULA 

In  each  of  the  following  find  the  term  called  for  without 
finding  any  other  term : 

23.  The  5th  term  of  (a  +  hy\ 

24.  The  7th  term  of  (3  a;  -  2  y)^, 

25.  The  6th  term  of  {Vx~  -Vyf^ 

26.  The9th  termof  (a;-?/)-'. 

27.  The  8th  term  of  ft  m  -  J  nf\ 

28.  The  7th  term  of  {a^h  -  ab^)^.      ' 

29.  The  6th  term  of  (a-a-^)^. 

30.  The  11th  term  of  {x^y-x'-y-y^. 

31.  The  5th  term  from  each  end  of  the  expansion  of  (a— &)^. 

32.  The  7th  term  from  each  end  of  (a Va  -  hVhfK 

33.  Which  term,  counting  from  the  beginning,  has  the  same 
coefficient  as  the  7th  term  of  (a  +  hf^  ?  Verify  by  finding  both 
coefficients.     How  do  the  exponents  differ  in  these  terms'? 

34.  What  other  term  has  the  same  coefficient  as  the  19th 
term  of  (a  +  hy^  ?  How  do  the  exponents  differ  ?  Find  in  the 
shortest  way  the  21st  term  of  (a  +  b)^. 

35.  ^  Find  the  87th  term  of  (a  +  b)^. 

36.  Find  the  53d  term  of  (a*  -  b^f^ 

37.  What  other  term  has  the  same  coefficient  as  the  5th 
term  in  the  expansion  of  (x  +  yy^  ? 

38.  Expand  [(a  +  6)  +  cf  by  the  binomial  formula. 

39.  Expand  [1  +  (2  a;  -f  3  y)y  by  the  binomial  formula. 

40.  Expand  (2x  —  Sy-{-4:zy  by  the  binomial  formula. 

41.  Write  the  Qc  +  l)st  term  of  (a  +  by.  Write  the  (n  +  l)st 
term  of  (a  +  by.  Show  that  the  next  and  also  all  succeeding 
terms  after  the  (n  -\-  l)st  term  have  zero  coefficients,  thus  prov- 
ing that  there  are  exactly  n-\-l  terras  in  the  expansion. 


INDEX 


[References  are  to  pages.] 


Abscissa 

Absolute  value 37, 

Addend  

Addition 5,39, 

axioms  of 

by  counting 

of  fractions   ....     83,  224, 

of  polynomials 57, 

principles  of t>,  57, 

of  radicals 186, 

of  signed  numbers      .     .     .38, 
Algebraic  expression   .     .     .31, 

fractions 215,  238, 

operations 4, 

sum 

Alternation 238, 

Antecedent 23G, 

Approach  as  a  limit  .... 
Approximate  roots    .     .      182, 
Area  problems  .     191,  192,  397, 

Arithmetic  means 

progression 

Associative  law      .    57,  67,  278, 
Averages  of  signed  numbers 
Axes  of  coordinates  .... 
Axioms,  of  addition   and   sub- 
traction   

of  multiplication  and  division 

Base  of  a  power      ....  134, 
Binomial 

expansion  of 

formula 

Brace,  bracket,  etc.  ...  7, 


109 
282 
278 
288 
277 

38 
402 
291 
282 
432 
282 
287 
398 
287 

39 
421 
421 
466 
362 
445 
460 
457 
280 

41 
109 

277 
279 

353 
5() 
474 
471 
293 


Characteristic  of  a  logarithm 
Checking  results  (>,  28,  149,  202, 


449 

etc. 

477 


Clearing  of  fractions      88,  245,  409 

Coefficient 5 

Common  denominator .     82,  220,  402 

difference 457 

factor 5,  215,  339 

multiple 82,  339 

Commutative  law  .  .  57,  278,  280 
Completing  the  square  .  199,  201 
Complex  fraction     .     .     .      233, 408 

number 351, 439 

Conditional  equation     20,  258,  302 

Consequent 236, 421 

Consistent  equations     .    .    .    322 

Constant 417 

Contradictory  equations  .     .     321 

Coordinates 109 

Cross  products 156 

Cube  root 287,  357,  360 

Cubes,  sum  or  difference  of 

152,  153,  329 

Degree  of  an  equation 

111,  128,166,365 
Dependent  equations     ...    321 

Determinants 319 

Difference 6,42,279 

of  two  cubes      ....      153,  329 
of  two  squares  ....      144,  329 

Digit  problems 65 

Directions  for  written  work .      25 


Distributive  law     .     .    11, 

280,  286 

Dividend,  divisor    .    .    . 

.  49,  281 

Divisi(^n,  axioms  of  .     .     . 

.     279 

of  fractions   ....     86, 

228,  405 

indicated 

.2,81 

by  means  of  logarithms 

.     45(5 

by  a  monomial       .     .    139, 

140,  290 

478 


INDEX 


[References 

Division  {Continued)  — 

by  a  polyuomial     .     .     .       148,  299 
principles  of     .    10,  13,  49,  138,  284 

of  a  product .13 

of  signed  numbers     ....      48 
of  a  sum  or  difference    ...      10 

by  zero 24, 149,  284 

Drill  Exercises  16,  55,  6(i,  80,  91, 
101,  115,  123,  133,  147,  165,  175, 
188,  198,  206,  214,  227,  235,  244, 
254 

Elimination,  by  addition  or  sub- 
traction   116, 316 

by  comparison 317 

by  formula 318 

by  substitution       .     .     .      118,  316 

Equality 7 

Equations 20 

conditional 20,  302 

contradictory 321 

consistent 322 

degree  of  .     .     .     Ill,  128,  166,  365 

equivalent 303 

graphic  representation  of  108,  365 
graphic  solution  of    .     .     .  112,  366 

identical 7,  20,  257 

independent 113,  321 

indeterminate   ....      314,  322 

irrational 196,  441 

linear Ill 

literal  .     1,  3,  92,  122,  258,  375,  etc. 
quadratic      .     .      166,  171,  199,  365 

roots  of 21,  302 

simultaneous     .      113,  116,  129,  316 

solution  of 24,  113,  128 

solved  by  factoring   .     .       166,  338 
systems  of     ...     .    113,  128,  207 

Exponent,  fractional    ....    424 

laws  of 136,  138,  424 

negative 139, 425 

of  a  power 134,  424 

zero 139,  425 

Expression,  algebraic,    4,  31,  56,  287 
integral,  fractional  81,  183,  215,  288 


are  to  pages.] 

mixed 

rational,  irrational     .     .      257, 
Extremes,  of  a  proportion     236, 

Factoring: 134,  261, 

equations  solved  by    .     .      166, 

Factor  theorem 

Factors 5,  134, 

found  by  the  factor  theorem  . 
found  by  grouping      .     .      158, 
highest  common     .     .     .      215, 
found  by  the  quadratic  for- 
mula       .      262, 

Form  changes         

Formulas    .     .    1,3,92,203,318, 

Fourth  proportional  .     .      240, 

Fractional,  equations  .     87,  245, 
exponents. 


expressions 
Fractions    . 

clearing  of 

complex    . 

reduction  of 
Fundamental  laws 

operations     .     . 


81,  183,  215, 
.  81, 215, 
.  88,  245, 
.     .      233, 

82,  219,  220, 
11,  57,  67, 


Geometric  means 
progressions  .     . 
Graphic  representation 

37,  102,  321, 
for  solution  of  equations 

108,  113,  ;%7, 
for  solution  of  problems     .     . 
Graphs,  of  linear  equations 

110,  260,  314, 
of  quadratic  equations   .     .     . 

of  statistics 

Greater  than 

Grouping-  terms  in  factoring 
158, 

Harmonic  means      ..... 

progressions 

Highest  common  factor    .  215, 
Historical  Notes  7, 11,19,  50,  57, 


288 
421 

329 
338 
335 
329 
335 
332 
339 

393 
23 

369 
421 
409 
424 

288 
398 
409 
408 
398 
277 
287 

465 
463 

365 

382 
106 

321 
365 

102 

282 

332 

469 
468 
339 
114 


INDEX 


479 


[References  are  to  pages.] 


Identities  .  .  •  •.  T.  22,  257,  302 
Imag-inary  unit  .  ,^^''i  ,351,  371,  439 
Incommensurable  ....  417 
Inconsistent  equations  .  .  .  321 
Indeterminate  equations    314,  321 

Index i;J4,287 

Induction,  mathematical .  .  .  472 
Integral  equations 303 

solutions 315 

Interest  problems  .  .  3,  29,  92,  467 
Inversion  of  a  proportion  2:38,  421 
Irrational,  number   .     .  193,  257,  350 

root 287 

Laws,  of  exponents  .     .136, 138,  424 

fundamental 277 

Less  than 282 

Lever  problems  ....  98,  271 
Like  terms  .  .'  .  .  .57, 58, 288 
Limit,  jtpproach  as  a  ....  466 
Linear  equations  .  .111,  116,  314 
and  quadratics  .  .  .  171,  207,  372 
Literal  equations 

1,  3,  92, 122,  258,  375,  etc 

fractions 215,  398 

problems.     .     .     .  93,  191,  394,  etc. 

Logarithms 449 

tal)le  of 452, 453 

Long  division 148 

Lo^vest  common  multiple 

82,221,340 
Lowest  terms  of  a  fraction  219,  398 

Mantissa  of  a  logarithm  .  .  449 
Mathematical  induction  .  .  472 
Mean  proportional      .     .     240, 421 

Means,  arithmetic 460 

geometric 465 

harmonic       469 

of  a  proportion       .     .     .      236,  421 
Members  of  an  equality    .    .  7,  24 

Minuend 42,  279 

Monomial  .  .  .  .50,  2S8,  294,  296 
Motion  problems 

95,  268,  275,  310,  326,  416,  462 


Multiple      82,  217,  339 

lowest  common  ....      217,  339 

Multiplication    .     .     .     .    9,  12, 294 

axioms  of .     . 279 

of  fractions  .  .  .  .  '  86,  228,  405 
by  means  of  logarithms  .  .  455 
of  polynomials       ....    67,  297 

of  radicals 194,  433 

of  signed  numbers  ...  46,  285 
principles  of  .  9,  12,  47,  70,  136,  284 
by  zero 284 

Negative,  exponent      .    .      139, 425 

number 36,  50,  282 

results,  interpretation  of    .     .      51 

Number,  absolute  value  of  .   37,  282 
constant,  variable  .     .     .     Ill,  417 

expression 4,  56,  287 

positive  and  negative  .  .  156,  282 
rational,  irrational     .  193,  257,  350 

real,  imaginary 351  ^  ^-^  ^ 

system  of  algebra  37,  255,  257,  350 
unknown  .     .   21,  116,  129,  302,  325 


Operand 

Operations,  fundamental, 
laws  of 


287 
287 
282 

non-unique         349 

order  of 15,288 

uniqueness  of    .     .     .    277,  279,  281 
Ordinates 109 

Parabola 365 

Parameter 375 

Parentheses 7,  293 

removal  of 62,  293 

Polynomials   .     .     .56,  291,  297,  299 
addition  and  subtraction  of    .      57 

cube  root  of       357 

multiplication  of 67 

square  root  of 178 

Positive  number     ....    36,  50 

Power 134,  349,  425 

Prime  factor 141,  218 

Principal  root 352 


480 


INDEX 


[References 

Principles,  eighteen  fundamen- 
tal, see  List  on  page  x. 
importance  of    .     .     .     .     .    17,67 
Principles   based    on    argu- 
mentation,    concerning 
addition  and  subtraction    282 
concerning  the  binomial  for- 
mula        472 

concerning  equivalent    equa- 
tions         303 

concerning  factors     .    .    .  335,  342 
concerning  fractions      .     .  286,  405 
concerning  fractional  and  neg- 
ative exponents        .     .     .    426 
concerning  multiplication  and 

division       284 

Problems,  applying  theory  of 

radicals 189,  445 

classified  by  topics  .    65,  73,  95,  264 
emphasizing  use  of  formulas 

93,  189,  259,  269,  309,  394, 
414 
involving    data    of   intrinsic 
interest 

33,  131,  191,  243,  266,  326 

solved  by  means  of  graphs     .    107 

Products,  of  fractions      86,  228,  405 

special 74,  134 

Progressions       457 

Proof  by  induction    ....    472 

Proportion 236,  421 

Pythagorean  proposition      .    172 

Quadratics  .  .  166,  171,  199,  365 
equations  in  the  form  of  .  .  389 
exposition  by  graphs  .  .  .  365 
properties  of  roots  of  .  .  202,  390 
systems  involving  171,  207,  372,  383 
solution   by    completing   the 

square  ....  199,  200,  201 
solution  by  factoring  .  .  166,  338 
solution  by  formula      203,  369,  390 

Quotients,  of  fractions  86,  228,  405 
of  polynomials  ....  148,  299 
special 152,  153 


are  to  pages.] 

Radical  expressions     185, 193,  429 

Radicals 176,  184,  429 

and  exponents 425 

applications  of  .  .  .  189,  193,  445 
equations  involving  .  .  .  196,  441 
simplification  of    ...     .  185, 193 

Radicand 429 

Ratio 236,  417 

Rational,  equation 303 

number 193,  257,  350 

root 287 

Rationalizing  the  divisor  195,  437 
Real  number  system     ...    350 

graph  of 37 

Reciprocal  of  a  number     .    .    405 

Remainder 42,  279 

Reviews    17,  35,  53,   79,  90,  100,  114, 

132,  164,  197,  213,  255 

Roots  and  powers      .     .  134,  142,  349 

cube 357 

square 142,  176,  358  . 

Roots  of  equations     .     .     .  21,  302 
distinct,  coincident,  imaginary,  369 

Scale  of  signed  numbers  .  .  37 
Signed  numbers  .  .  .  .  37, 45 
Signs,  of  aggregation    ...     7,  293 

of  inequality 282 

of  operation 2 

of  quality 37 

Similar,  radicals  ....      186,  432 

terms 57,  288 

triangles 241 

Simultaneous  equations 

113,  116,  128,  322 

fractional 248 

higher 388 

linear 113,116,314 

linear  and  quadratic     171,  207,  372 

quadratic       . 379 

Solution  of  equations 

20,  113,  261,  314 

by  formula 203,369 

by  graphs  ....  113,  322,  365 
by  special  devices      ....    378 


INDEX 


481 


[References 

Solution  of  problems,  classi- 
fied list  93,264 

hints  on 27,  31 

Square  of  binomial    ....      74 
Square  root    ....   142,  159,  176 

applications  of 189,  445 

approximate 182 

of  a  radical  expression  .     .     .    438 

of  fractions 183 

of  monomials 177 

of  polynomials 178 

Subtraction 5,  42,  279 

axioms  of . 277 

effractions   ....     83,224,402 

of  polynomials 67,  291 

of  radicals 186,  432 

of  signed  numbers     ...  42,  282 
principles  of      ....    6,  43,  282 

Substitution 301 

elimination  by 118,  316 

Sum 6,  38, 186,  277 

Surd 193 

Symbols,  of  aggregation  .     .     7,  293 

of  operation 3 

Systems  of  equations  113,  116, 128, 
171,  207,  261,  314,  323,  372 


Table  of  logarithms 
Tangrent  to  a  curve 


452 
370 


are  to  pages.] 

Terms       288 

compound 56 

of  a  fraction      ...     81,  219,  399 
of  a  proportion  ....      236,  421 

simple 56 

Third  proportional     ....    421 

Trinomial 56 

Trinomial  squares      .    .      142,  329 

Unknown  numbers 

21,  124,  128,  260,  302,  375 

Uniqueness,  of  addition  ...  277 

of  division 281 

of  multiplication 279 

of  subtraction 279 

Unity,  definition  of 281 

Variables    .    .    21,124,128,260,417 

Variation 417 

direct 236,418 

inverse 418 

Vinculum 294 

Written  work,  directions  for  .      25 

Zero,  definition  of 279 

division  by  .    24,  149,  285,  304,  349 
multiplication  by  .     .    24,  284,  304 


MATHEMATICS 


First  Principles  of  Algebra 

By  H.  E.  Slaught,  Associate  Professor  of  Mathematics  in  the  Univer- 
sity of  Chicago,  and  N.  J.  Lenne:s,  Instructor  in  Mathematics  in 
Columbia  University,  New  York  City.  Elementary  Course,  i2mo, 
cloth,  288  pages.  Price,  $1.00.  ADVANCED  COURSE  and  COM- 
PLETE Course,  Ai  Press. 

THIS  book  embodies  the  methods  of  what  might  be  called  the 
new  school  of  Algebra  teaching,  but  at  the  same  time  has 
kept  the  valuable  features  of  the  books  which  preceded  this 
movement. 

In  writing  the  First  Principles  of  Algebra  the  authors  have  been 
governed  by  two  main  purposes:  (i)  to  provide  a  gradual  and 
natural  introduction  to  the  symbols  and  processes  of  algebra; 
(2)  to  give  purpose  to  the  subject  of  algebra  by  a  constant  use  of 
it  in  doing  interesting  and  valuable  things.  Each  of  these  pur- 
poses leads  to  the  same  order  of  topics,  which,  however,  differs  in 
minor  features  from  the  conventional  order.  In  the  Introduction 
the  authors  offer  a  full  explanation  of  their  reasons  for  adopting 
the  present  order,  which  will  be  found  to  be  logical  and 
systematic. 

The  study  of  equations  and  their  uses  is  regarded  as  the  main 
topic  for  the  first  year's  work.  It  is  recognized  that  abstract  equa- 
tions will  appear  of  little  or  no  value  to  the  pupil  unless  he  finds 
uses  for  them  ;  hence  frequent  lists  of  problems  are  provided  for 
translation  into  equations  and  for  solution.  Many  of  these  prob- 
lems involve  valuable  mathematical  concepts,  so  that  during  the 
first  half  year  algebra  is  made  to  appeal  to  the  higher  and  more 
useful  types  of  interest,  and  not  merely  to  the  instinct  for  solv- 
ing puzzles,  which  must  be  the  case  if  the  greater  part  of  this  time 
is  spent  on  factoring  and  in  manipulating  complicated  fractions. 

The  principles  of  algebra  used  in  the  Elementary  Course  are 
stated  in  a  small  number  of  short  rules,  eighteen  in  all.  The 
purpose  of  these  rules  is  to  furnish  in  simple  form  a  codifi- 
cation of  those  operations  of  algebra  which  require  special  em- 
phasis. 

70 


MATHEMATICS 


Plane  Geometry  with  Problems  and  Applications 

By  H.  E.  Slaught,  Associate  Professor  of  Mathematics  in  the  Uni- 
versity of  Chicago,  and  N.  J.  Lennes,  Instructor  in  Mathematics  in 
Columbia  University,  New  York  City.  i2mo,  cloth,  288  pages.  Price, 
^i.oo. 

THE  two  main  objects  that  the  authors  have  had  in  view  in 
preparing  this  book  are  to  develop  in  the  pupil,  gradually 
and  imperceptibly,  the  power  and  the  habit  of  deductive  reasoning, 
and  to  teach  the  pupils  to  recognize  the  essential  facts  of  elemen- 
tary geometry  as  properties  of  the  space  in  which  they  live, 
not  merely  as  statements  in  a  book. 

These  two  objects  the  book  seeks  to  accomplish  in  the  follow- 
ing ways :  — 

1 .  The  simplification  of  the  first  five  chapters  by  the  exclusion 
of  some  theorems  found  in  current  books. 

2.  By  introducing  many  applications  of  special  interest  to 
pupils,  and  by  including  only  such  concrete  problems  as  fairly 
come  within  the  knowledge  of  the  average  pupil.  Such  problems 
may  be  found  in  tile  patterns,  parquet  floors,  linoleums,  wall 
papers,  steel  ceilings,  grill  work,  ornamental  windows,  etc.,  and 
they  furnish  a  large  variety  of  simple  exercises  both  for  geometric 
construction  and  proofs  and  for  algebraic  computation. 

3.  The  pupil  approaches  the  formal  logic  of  geometry  by 
natural  and  gradual  processes.  At  the  outset  the  treatment  is 
informal ;  the  more  formal  development  that  follows  guides  the 
student  by  questions,  outlines,  and  other  devices,  into  an  attitude 
of  mental  independence  and  an  appreciation  of  clear  reasoning. 

The  arrangement  of  the  text  is  adapted  to  three  grades  of 
courses :  — 

(a)  A  minimum  course,  providing  as  much  material  as  is  found 
in  the  briefest  books  now  in  use. 

(d)  A  medium  course,  fully  covering  the  entrance  require- 
ments of  any  college  or  technical  school. 

(c)  An  extended  course,  furnishing  ample  work  for  those 
sch6ols  where  the  students  are  more  mature,  or  where  more  time 
can  be  given  to  the  subject. 

72 


MATHEMATICS 


Introduction  to  Geometry 

By  William  Schoch,  of  the  Crane  Manual  Training  High  School 
Chicago.     i2mo,  cloth,  142  pages.     Price,  60  cents. 

THIS  book  is  intended  for  pupils  in  the  upper  grades  of 
the  grammar  schools.  It  is  a  book  of  exercises  and 
problems  by  means  of  which  the  pupil  will  convince  himself 
of  a  number  of  fundamental  geometric  truths.  Practical 
problems  are  given  and  the  pupil's  knowledge  is  constantly 
tested  by  his  power  to  apply  it. 

Principles  of  Plane  Geometry 

By  J.  W.  MacDonald,  Agent  of  the  Massachusetts  Board  of  Educa- 
tion.    i6mo,  paper,  70  pages.     Price,  30  cents. 

Logarithmic  and  Other  Mathematical  Tables 

By  William  J.  Hussey,  Professor  of  Astronomy  in  the  Leland  Stan- 
ford Junior  University,  California.    8vo,  cloth,  148  pages.     Price,  ;^i.oo. 

VARIOUS  mechanical  devices  make  this  work  specially  easy 
to  consult;  and  the  large,  clear,  open  page  enables  one 
readily  to  find  the  numbers  sought.  It  commends  itself  at  once 
to  the  eye,  as  a  piece  of  careful  and  successful  book-making. 

Plane  Trigonometry 

By  Professor  R.  D.  BoHANNAN,  of  the  State  University,  Columbus, 
Ohio.    Cloth,  379  pages.    Price,  jg2.sa 

Calculus  with  Applications 

By  Ellen  Hayes,  Professor  of  Mathematics  at  Wellesley  College. 
i2mo,  cloth,  170  pages.    Price,  ^1.20. 

THIS  book  is  a  reading  lesson  in  applied  mathematics,  in- 
tended for  persons  who  wish,  without  taking  long  courses  in 
xnathematics,  to  know  what  the  calculus  is  and  how  to  use  it,  either 
as  applied  to  other  sciences,  or  for  purposes  of  general  culture. 

76 


HISTORY 


Readings  in  Ancient  History:  A  Selection  of  Illus- 
trative Extracts  from  the  Sources 

By  Professor  WILLIAM  STEARNS  DAVIS,  of  the  University  of  Minne- 
sota; Introduction  by  Professor  WILLIS  MASON  WEST,  of  the  Univer- 
sity of  Minnesota. 

Volume  I :  Greece  and  the  East.     i2mo,  cloth,  ooo  pages.    Price,  ^o.oo. 

Volume  II :  Rome  and  the  West.     i2mo,  cloth,  coo  pages.    Price,  ^o.oo. 

THIS  book  sets  before  the  student  beginning  the  study  of 
Ancient  History  a  sufficient  amount  of  source  material  to 
fllnstrate  the  important  or  typical  historical  facts  which  will  be 
mentioned  in  his  text-book.  The  volumes  are  not  designed  for 
hard  study,  to  be  tested  scrupulously  by  minute  questioning; 
they  are  meant  for  reading,  —  a  daily  companion  to  any  standard 
text  in  Ancient  History,  —  and  the  boy  or  girl  so  using  them  is 
sure  to  breathe  in  more  of  the  atmosphere  of  the  ancient  world, 
and  to  get  more  taste  of  the  notable  literary  flavor  pervading 
Greek  and  Roman  history,  than  would  be  possible  from  the  study 
of  a  conventional  text-book. 

Volume  I  contains  125  different  selections,  of  which  the  follow^- 
ing  are  typical :  The  Ethics  of  an  Egyptian  Nobleman,  Inscrip- 
tion ;  An  Assyrian  Palace,  Maspero ;  The  Shield  of  Achilles,  T/ie 
Iliad',  How  Glaucus  tried  to  tempt  the  Delphic  Oracle,  Herodo- 
tus ;  The  Ring  of  Polycrates,  Herodotus ;  How  Leonidas  held  the 
Pass  of  Thermopylae,  Herodotus;  The  Last  Fight  in  the  Harbor 
of  Syracuse,  Thucydides ;  Anecdotes  about  Socrates,  Diogenes 
Laertius ;  How  Lysias  escaped  from  the  "Thirty,"  Lysias ;  How 
Elephants  fought  in  Hellenistic  Armies,  Polybius. 

Volume  H  contains  148  selections,  including :  Brutus  condemns 
his  own  Sons  to  Death,  Livy  ]  How  the  Plebeians  won  the  Con- 
sulship, Livy ;  The  Honesty  of  Roman  Officials,  Polybius ;  The 
Reign  of  Terror  under  Sulla,  Plutarch ;  The  Wealth  and  Habits 
of  Crassus  the  Millionaire,  Plutarch;  The  Personal  Traits  of 
Julius  Caesar,  Suetonius;  A  Business  Panic  in  Rome,  Tacitus; 
The  Bill  of  Fare  of  a  Great  Roman  Banquet,  Macrobius ;  How  a 
Stoic  met  Calamity  in  the  Days  of  Nero,  Epictetus ;  The  Precepts 
of  Marcus  Aurelius,  Marcus  Aurelius. 

78 


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OCT  24  1946 


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THE  UNIVERSITY  OF  CALIFORNIA  LIBRARY 


